Module THCS 18

PHNG PHP DY HC TCH CC | 57

MODULE THcs

18

PHng php dy hc

tch cc

TRN NH CHU NG TH THU THU

PHAN TH LUYN

58 | MODULE THCS 18

A. GII THIU TNG QUAN

Sucthnang pht trin kinh t x hi trong bi cnh ton cu ho t ra nhucthngang

yu cu m"i i v"i ngucth%i lao ng, do cuthngang t ra nhucthngang yu cu

m"i cho sucthnang nghi,p gio duthnangc th h, tr. v o t/o ngu0n nhn lucthnangc. Gio

duthnangc cn o t/o i nguthnga nhn lucthnangc c kh n4ng p ucthsacng ucth6c nhucthngang i

h8i m"i cuthhoia x hi v th: tructh%ng lao ng, c bi,t l n4ng lucthnangc hnh

ng, tnh n4ng ng, sng t/o, tnh tucthnang lucthnangc v trch nhi,m cuthngang nhucth

n4ng lucthnangc cng tc lm vi,c, n4ng lucthnangc gii quyt cc v=n > phucthsacc h6p.

?@i m"i PPDH l mt trong nhucthngang nhi,m vuthnang quan trDng cuthhoia @i m"i

gio duthnangc, ucth6c nu v thucthnangc hi,n t nh=t l trong vi chuthnangc n4m nay E

mDi tructh%ng ph@ thng trn c nucth"c. V> nguyn tHc, c th xem vi,c @i

m"i PPDH ucth6c bHt u thucthnangc hi,n tucthhuyen sau ?/i hi ln thucthsac VI cuthhoia ?ng

Cng sn Vi,t Nam. Tuy nhin, @i m"i PPDH thucthnangc sucthnang trE thnh mt ho/t

ng rng khHp trong ton ngnh tucthhuyen sau vi,c ban hnh Ngh: quyt 4 cuthhoia

Ban Ch=p hnh Trung ucthQng ?ng cng sn kho VII v"i yu cu tip tuthnangc

@i m"i muthnangc tiu, ni dung, chucthQng trnh, phucthQng php gio duthnangc.

Ngh: quyt v> gio duthnangc v khoa hDc cng ngh, cuthhoia Hi ngh: ln thucthsac hai

Ban ch=p hnh Trung ucthQng ?ng kho VIII tip tuthnangc nh=n m/nh v cuthnang th

ho hQn yu cu @i m"i PPDH. Tucthhuyen n nay, phucthQng php gio duthnangc,

PPDH lun lun ucth6c > cVp khi nh gi gio duthnangc trong cc v4n ki,n

cuthhoia ?ng v Nh nucth"c. Trong th%i gian qua, mc du c nhucthngang nW lucthnangc

@i m"i PPDH ng ghi nhVn trong ton ngnh, tructh"c ht l gio duthnangc ph@

thng nhucthng Bo co Chnh tr: cuthhoia Ban Ch=p hnh Trung ucthQng ?ng

Cng sn Vi,t Nam ln thucthsac XI vYn tip tuthnangc nhVn :nh: chucthQng trnh,

ni dung, phucthng php dy v hc lc hu, i mi chm. Ngh: quyt

?/i hi ?ng ln ny t ra yu cu i mi cn bn v ton din n>n

gio duthnangc nucth"c nh, mt nhi,m vuthnang ht sucthsacc l"n lao cho ton ngnh Gio

duthnangc nucth"c ta, trong c vi,c tip tuthnangc [y m/nh @i m"i PPDH.

?:nh hucth"ng quan trDng trong @i m"i PPDH l pht huy tnh tch cucthnangc, tucthnang

lucthnangc v sng t/o, pht trin n4ng lucthnangc hnh ng, n4ng lucthnangc cng tc lm

vi,c cuthhoia ngucth%i hDc. ? cuthngang l nhucthngang xu hucth"ng quc t trong ci cch

PPDH E nh tructh%ng ph@ thng.


? thucthnangc hi,n c hi,u qu vi,c @i m"i PPDH E tructh%ng ph@ thng vi,c

o t/o v b0i ducth\ng i nguthnga GV c n4ng lucthnangc d/y hDc theo nhucthngang quan

im @i m"i PPDH c vai tr then cht. Tucthhuyen nhi>u n4m nay, B Gio duthnangc

v ?o t/o ch vi,c b0i ducth\ng GV v> @i m"i PPDH v c nhi>u

ti li,u v> chuthhoi > ny ucth6c xu=t bn. Module ny trnh by mt s cQ sE

thucthnangc ti`n v l luVn chung, cuthngang nhucth mt s quan im, phucthQng php v

ka thuVt d/y hDc tch cucthnangc c th p duthnangng trong vi,c @i m"i PPDH, nhbm

gip GV c ci nhn t@ng quan v> @i m"i PPDH, trn cQ sE c th tm

ucth6c nhucthngang tucthEng, g6i vVn duthnangng vo cc mn hDc cuthnang th. Module

khng c tham vDng trnh by ton di,n v> chuthhoi > ny, m chc tVp trung

vo mt s v=n > lucthnanga chDn. Trong mWi v=n > chc trnh by nhucthngang ni

dung cQ bn, lm cQ sE cho vi,c vVn duthnangng cuthngang nhucth cho vi,c tm hiu,

tho luVn tip theo.

B. MC TIU

Sau khi hDc xong module ny, hDc vin cn:

Tm tHt ucth6c :nh hucth"ng @i m"i PPDH.

Li,t k cc c tructhng cuthhoia PPDH tch cucthnangc.

Nu ucth6c mt s PPDH tch cucthnangc.

Tm tHt ucth6c bn ch=t, quy trnh, ucthu, nhucth6c im cuthhoia mWi PPDH ucth6c

gi"i thi,u trong module ny.

VVn duthnangng ucth6c cc PPDH tch cucthnangc vo chuyn mn cuthhoia mnh mt cch

linh ho/t, sng t/o

?eI TuchoagNG Suchoahoi DuhoanangNG

GV, cn b chc /o chuyn mn, cn b qun l c=p THCS.

60 | MODULE THCS 18

C. NI DUNG

Ni dung 1

TM HIU V PHNG PHP DY HC TCH CC

V CC C TRNG CA PHNG PHP DY HC TCH CC

NHIM V

B/n hy Dc v nghin cucthsacu nhucthngang thng tin cQ bn phn tch, lm r:

1. PhucthQng php d/y hDc tch cucthnangc l g? Bn ch=t cuthhoia phucthQng php d/y hDc

tch cucthnangc nhucth th no?

2. Nhucthngang c tructhng cQ bn cuthhoia phucthQng php d/y hDc tch cucthnangc.

THNG TIN C BN

1. Phng php dy hc tch cc

?:nh hucth"ng @i m"i phucthQng php d/y v hDc ucth6c xc :nh trong

Ngh: quyt Trung ucthQng 4 kho VII (1/1993), Ngh: quyt Trung ucthQng 2

kho VIII (12/1996), ucth6c th ch ho trong LuVt Gio duthnangc (02/12/1998),

ucth6c cuthnang th ho trong cc chc th: cuthhoia B Gio duthnangc v ?o t/o, c bi,t l

Chc th: s 15 (4/1999).

?i>u 28.2 cuthhoia LuVt Gio duthnangc (14/6/2005) ghi: Phucthng php gio duthnangc

ph thng phi pht huy tnh tch cucthnangc, tucthnang gic, chuthhoi &ng, sng to cuthhoia HS;

ph h-p vi .c i/m cuthhoia tucthhuyenng lp hc, mn hc; b1i ducth2ng phucthng php

tucthnang hc, rn luyn k6 nng vn duthnangng ki7n thucthsacc vo thucthnangc ti9n; tc &ng 7n

tnh cm, em li ni


khng phi tVp trung vo pht huy tnh tch cucthnangc cuthhoia ngucth%i d/y; tuy nhin,

d/y hDc theo phucthQng php tch cucthnangc th GV phi nW lucthnangc nhi>u so v"i

d/y hDc thuthnang ng.

2. c trng ca phng php dy hc tch cc

a. Dy hc thng qua t chucthsacc cc hot &ng hc tp cuthhoia HS

Trong PPDH tch cucthnangc, ngucth%i hDc i tucth6ng cuthhoia ho/t ng d/y, 0ng

th%i l chuthhoi th cuthhoia ho/t ng hDc ucth6c cun ht vo cc ho/t ng

hDc tVp do GV t@ chucthsacc v chc /o, thng qua tucthnang lucthnangc khm ph nhucthngang

i>u mnh chuctha r chucthsac khng phi thuthnang ng tip thu nhucthngang tri thucthsacc

ucth6c GV sHp t. ?ucth6c t vo nhucthngang tnh hung cuthhoia %i sng thucthnangc t,

ngucth%i hDc tructhnangc tip quan st, tho luVn, lm th nghi,m, gii quyt v=n >

t ra theo cch suy ngha cuthhoia mnh, tucthhuyen nHm ucth6c kin thucthsacc ka n4ng

m"i, vucthhuyena nHm ucth6c phucthQng php lm ra kin thucthsacc, ka n4ng , khng

rVp theo nhucthngang khun mYu syn c, ucth6c bc l v pht huy ti>m n4ng

sng t/o.

D/y theo cch ny, GV khng chc gin Qn truy>n /t tri thucthsacc m cn hucth"ng

dYn hnh ng. ChucthQng trnh d/y hDc phi gip cho tucthhuyenng HS bit hnh

ng v tch cucthnangc tham gia cc chucthQng trnh hnh ng cuthhoia cng 0ng.

b. Dy hc ch trng rn luyn phucthng php tucthnang hc

PhucthQng php tch cucthnangc xem vi,c rn luy,n phucthQng php hDc tVp cho HS

khng chc l mt bi,n php nng cao hi,u qu d/y hDc m cn l mt

muthnangc tiu d/y hDc.

Trong x hi hi,n /i ang bin @i nhanh v"i sucthnang bng n@ thng tin,

khoa hDc, ka thuVt, cng ngh, pht trin nhucth vuthnga bo th khng th nh0i

nht vo u c HS khi lucth6ng kin thucthsacc ngy cng nhi>u. Phi quan tm

d/y cho HS phucthQng php hDc ngay tucthhuyen bVc Tiu hDc v cng ln bVc hDc

cao hQn cng phi ucth6c ch trDng.

Trong cc phucthQng php hDc th ct li l phucthQng php tucthnang hDc. Nu rn

luy,n cho ngucth%i hDc c ucth6c phucthQng php, ka n4ng, thi quen, ch tucthnang

hDc th s| t/o cho hD lng ham hDc, khQi dVy ni lucthnangc vn c trong mWi

con ngucth%i, kt qu hDc tVp s| ucth6c nhn ln g=p bi. V vVy, ngy nay

62 | MODULE THCS 18

ngucth%i ta nh=n m/nh mt ho/t ng hDc trong qu trnh d/y hDc, nW lucthnangc

t/o ra sucthnang chuyn bin tucthhuyen hDc tVp thuthnang ng sang tucthnang hDc chuthhoi ng, t v=n

> pht trin tucthnang hDc ngay trong tructh%ng ph@ thng, khng chc tucthnang hDc E nh

sau bi ln l"p m tucthnang hDc c trong tit hDc c sucthnang hucth"ng dYn cuthhoia GV.

c. Tng cucth@ng hc tp c th/, phDi h-p vi hc tp h-p tc

Trong mt l"p hDc, trnh kin thucthsacc, tucth duy cuthhoia HS khng th 0ng

>u tuy,t i nn khi p duthnangng phucthQng php tch cucthnangc buc GV v HS phi

ch=p nhVn sucthnang phn ho v> cucth%ng , tin hon thnh nhi,m vuthnang hDc

tVp, nh=t l khi bi hDc ucth6c thit k thnh mt chuWi cng tc c lVp.

p duthnangng phucthQng php tch cucthnangc E trnh cng cao th sucthnang phn ho ny

cng l"n. Vi,c sucthhoi duthnangng cc phucthQng ti,n CNTT trong nh tructh%ng s| p

ucthsacng yu cu c th ho ho/t ng hDc tVp theo nhu cu v kh n4ng cuthhoia

mWi HS.

Tuy nhin, trong hDc tVp, khng phi mDi tri thucthsacc, ka n4ng, thi >u

ucth6c hnh thnh bbng nhucthngang ho/t ng c lVp c nhn. L"p hDc l mi

tructh%ng giao tip thy tr, tr tr, t/o nn mi quan h, h6p tc giucthngaa

cc c nhn trn con ucth%ng chim lanh ni dung hDc tVp. Thng qua

tho luVn, tranh luVn trong tVp th, kin mWi c nhn ucth6c bc l,

khng :nh hay bc b8, qua ngucth%i hDc nng mnh ln mt trnh

m"i. Bi hDc vVn duthnangng ucth6c vn hiu bit v kinh nghi,m sng cuthhoia

ngucth%i thy gio.

Trong nh tructh%ng, phucthQng php hDc tVp h6p tc ucth6c t@ chucthsacc E c=p

nhm, t@, l"p hoc tructh%ng. ?ucth6c sucthhoi duthnangng ph@ bin trong d/y hDc l ho/t

ng h6p tc trong nhm nh8 4 n 6 ngucth%i. HDc tVp h6p tc lm t4ng

hi,u qu hDc tVp, nh=t l lc phi gii quyt nhucthngang v=n > gay c=n, lc

xu=t hi,n thucthnangc sucthnang nhu cu phi h6p giucthngaa cc c nhn hon thnh

nhi,m vuthnang chung. Trong ho/t ng theo nhm nh8 s| khng th c hi,n

tucth6ng l/i; tnh cch, n4ng lucthnangc cuthhoia mWi thnh vin ucth6c bc l, un nHn,

pht trin tnh b/n, thucthsacc t@ chucthsacc, tinh thn tucthQng tr6. M hnh h6p tc

trong x hi uctha vo %i sng hDc ucth%ng s| lm cho cc thnh vin quen

dn v"i sucthnang phn cng h6p tc trong lao ng x hi.


Trong n>n kinh t th: tructh%ng xu=t hi,n nhu cu h6p tc xuyn quc

gia, lin quc gia; n4ng lucthnangc h6p tc phi trE thnh mt muthnangc tiu gio duthnangc

m nh tructh%ng phi chu[n b: cho HS.

d. K7t h-p nh gi cuthhoia thFy vi tucthnang nh gi cuthhoia tr

Trong d/y hDc, vi,c nh gi HS khng chc nhbm muthnangc ch nhVn :nh

thucthnangc tr/ng v i>u chcnh ho/t ng hDc cuthhoia tr m cn 0ng th%i t/o

i>u ki,n nhVn :nh thucthnangc tr/ng v i>u chcnh ho/t ng d/y cuthhoia thy.

Tructh"c y GV giucthnga c quy>n nh gi HS. Trong phucthQng php tch cucthnangc,

GV phi hucth"ng dYn HS pht trin ka n4ng tucthnang nh gi tucthnang i>u chcnh

cch hDc. Lin quan v"i i>u ny, GV cn t/o i>u ki,n thuVn l6i HS

ucth6c tham gia nh gi lYn nhau. Tucthnang nh gi ng v i>u chcnh ho/t

ng k:p th%i l n4ng lucthnangc r=t cn cho sucthnang thnh /t trong cuc sng m

nh tructh%ng phi trang b: cho HS.

Theo hucth"ng pht trin cc phucthQng php tch cucthnangc o t/o nhucthngang con

ngucth%i n4ng ng, s"m thch nghi v"i %i sng x hi th vi,c kim tra,

nh gi khng th ducthhuyenng l/i E yu cu ti hi,n cc kin thucthsacc, lp l/i cc ka

n4ng hDc m phi khuyn khch tr thng minh, c sng t/o trong vi,c

gii quyt nhucthngang tnh hung thucthnangc t.

V"i sucthnang tr6 gip cuthhoia cc thit b: ka thuVt, kim tra, nh gi s| khng cn

l mt cng vi,c nng nhDc i v"i GV, m l/i cho nhi>u thng tin k:p

th%i hQn linh ho/t i>u chcnh ho/t ng d/y, chc /o ho/t ng hDc.

Tucthhuyen d/y v hDc thuthnang ng sang d/y v hDc tch cucthnangc, GV khng cn ng vai

tr Qn thun l ngucth%i truy>n /t kin thucthsacc, m trE thnh ngucth%i thit k,

t@ chucthsacc, hucth"ng dYn cc ho/t ng c lVp hoc theo nhm nh8 HS tucthnang

lucthnangc chim lanh ni dung hDc tVp, chuthhoi ng /t cc muthnangc tiu kin thucthsacc, ka

n4ng, thi theo yu cu cuthhoia chucthQng trnh. Trn l"p, HS ho/t ng l

chnh, GV c v. nhn nh hQn. Nhucthng khi so/n gio n, GV phi u tucth

cng sucthsacc, th%i gian r=t nhi>u so v"i kiu d/y v hDc thuthnang ng m"i c th

thucthnangc hi,n bi ln l"p v"i vai tr l ngucth%i g6i mE, xc tc, ng vin, c

v=n, trDng ti trong cc ho/t ng tm ti ho hucthsacng, tranh luVn si n@i

cuthhoia HS. GV phi c trnh chuyn mn su rng, c trnh sucth ph/m

lnh ngh> m"i c th t@ chucthsacc, hucth"ng dYn cc ho/t ng cuthhoia HS m nhi>u

khi di`n bin ngoi tm ducthnang kin cuthhoia GV.

64 | MODULE THCS 18

Ni dung 2

TM HIU V PHNG PHP DY HC GI M VN P

Hot ng 1. Tm hiu v phng php dy hc gi m vn p

NHIM V

B/n hy Dc v nghin cucthsacu nhucthngang thng tin cQ bn cuthhoia ho/t ng 1

lm r:

1. Bn ch=t cuthhoia PPDH g6i mE v=n p v quy trnh thucthnangc hi,n n.

2. Chc ra nhucthngang ucthu im, nhucthngang h/n ch v nhucthngang im cn lucthu v>

phucthQng php d/y hDc ny.

3. L=y v duthnang minh ho/.

THNG TIN C BN

PhucthQng php ny khEi thu tucthhuyen cch d/y hDc cuthhoia Xcrat. ?y l mt

PPDH thucth%ng xuyn ucth6c vVn duthnangng trong d/y hDc cc mn hDc E tructh%ng

THCS.

1. Bn cht ca PPDH gi m, vn p

PhucthQng php v=n p l qu trnh tucthQng tc giucthngaa GV v HS, ucth6c thucthnangc

hi,n thng qua h, thng cu h8i v cu tr l%i tucthQng ucthsacng v> mt chuthhoi >

nh=t :nh ucth6c GV t ra. Qua vi,c tr l%i h, thng cu h8i dYn dHt cuthhoia

GV, HS th hi,n ucth6c suy ngha, tucthEng cuthhoia mnh, tucthhuyen khm ph v lanh

hi ucth6c i tucth6ng hDc tVp.

?y l PPDH m GV khng tructhnangc tip uctha ra nhucthngang kin thucthsacc hon chcnh

m hucth"ng dYn HS tucth duy tucthhuyenng bucth"c cc em tucthnang tm ra kin thucthsacc m"i

phi hDc. C4n cucthsac vo tnh ch=t ho/t ng nhVn thucthsacc cuthhoia HS, ngucth%i ta

phn bi,t cc lo/i: v=n p ti hi,n, v=n p gii thch minh ho/ v v=n

p tm ti.

VIn p ti hin: ucth6c thucthnangc hi,n khi nhucthngang cu h8i do GV t ra chc yu

cu HS nhHc l/i kin thucthsacc bit v tr l%i ducthnanga vo tr nh", khng cn suy

luVn. V=n p ti hi,n c ngu0n gc tucthhuyen kiu d/y hDc gio i>u. L luVn

d/y hDc hi,n /i khng xem v=n p ti hi,n l mt phucthQng php c


gi tr: sucth ph/m. Lo/i v=n p ny chc nn sucthhoi duthnangng h/n ch khi cn t

mi lin h, giucthngaa kin thucthsacc hDc v"i kin thucthsacc sHp hDc hoc khi cuthhoing c

kin thucthsacc vucthhuyena m"i hDc.

VIn p gii thch minh ho ucth6c thucthnangc hi,n khi nhucthngang cu h8i cuthhoia GV

uctha ra c km theo cc v duthnang minh ho/ (bbng l%i hoc bbng hnh nh tructhnangc

quan) nhbm gip HS d` hiu, d` ghi nh". Vi,c p duthnangng phucthQng php ny

c gi tr: sucth ph/m cao hQn nhucthng kh hQn v i h8i nhi>u cng sucthsacc cuthhoia

GV hQn khi chu[n b: h, thng cc cu h8i thch h6p. PhucthQng php ny

ucth6c p duthnangng c hi,u qu trong mt s tructh%ng h6p, nhucth khi GV biu di` n

phucthQng ti,n tructhnangc quan.

VIn p tm ti (hay v=n p pht hi,n): l lo/i v=n p m GV t@ chucthsacc

sucthnang trao @i kin k c tranh luVn giucthngaa thy v"i c l"p, c khi giucthngaa tr

v"i tr, thng qua , HS nHm ucth6c tri thucthsacc m"i. H, thng cu h8i ucth6c

sHp t h6p l nhbm pht hi,n, t ra v gii quyt mt v=n > xc :nh,

buc HS phi lin tuthnangc c gHng, tm ti l%i gii p.

Trong v=n p tm ti, h, thng cu h8i cuthhoia GV giucthnga vai tr chc /o, quyt

:nh ch=t lucth6ng lanh hi cuthhoia l"p hDc. TrVt tucthnang logic cuthhoia cc cu h8i hucth"ng

dYn HS tucthhuyenng bucth"c pht hi,n ra bn ch=t cuthhoia sucthnang vVt, quy luVt cuthhoia hi,n

tucth6ng, kch thch tnh tch cucthnangc tm ti, sucthnang ham mun hiu bit cuthhoia HS.

2. Quy trnh thc hin

Tructhc gi hc:

Bucthc 1: Xc :nh muthnangc tiu bi hDc v Di tucth-ng d/y hDc. Xc :nh cc Qn

v: kin thucthsacc, ka n4ng cQ bn trong bi hDc v tm cch di`n /t cc ni

dung ny ducth"i d/ng cu h8i g6i , dYn dHt HS.

Bucthc 2: Ducthnang kin ni dung cc cu h8i, hnh thucthsacc h8i, th%i im t cu

h8i (t cu h8i E chW no?), trnh tucthnang cuthhoia cc cu h8i (cu h8i tructh"c phi

lm n>n cho cc cu h8i tip sau hoc :nh hucth"ng suy ngha HS gii

quyt v=n >). Ducthnang kin ni dung cc cu tr l@i cuthhoia HS, trong ducthnang kin

nhucthngang lW h@ng v> mt kin thucthsacc cuthngang nhucth nhucthngang kh kh4n, sai lm

ph@ bin m HS thucth%ng mHc phi. Ducthnang kin cc cu nhVn xt hoc tr l%i

cuthhoia GV i v"i HS.

66 | MODULE THCS 18

Bucthc 3: Ducthnang kin nhucthngang cu h8i phuthnang tu tnh hnh tucthhuyenng i tucth6ng cuthnang

th m tip tuthnangc g6i , dYn dHt HS.

Trong gi hc

Bucthc 4: GV sucthhoi duthnangng h, thng cu h8i ducthnang kin (ph h6p v"i trnh nhVn

thucthsacc cuthhoia tucthhuyenng lo/i i tucth6ng HS) trong tin trnh bi d/y v ch thu

thVp thng tin phn h1i tucthhuyen pha HS.

Sau gi hc

GV ch rt kinh nghi,m v> tnh r rng, chnh xc v trVt tucthnang logic cuthhoia

h, thng cu h8i ucth6c sucthhoi duthnangng trong gi% d/y.

3. u im

V=n p l cch thucthsacc tt kch thch tucth duy c lVp cuthhoia HS, d/y HS

cch tucthnang suy ngha ng Hn. Bbng cch ny, HS hiu ni dung hDc tVp hQn

l hDc vt, hDc thuc lng.

G6i mE v=n p gip li cun HS tham gia vo bi hDc, lm cho khng

kh l"p hDc si n@i, sinh ng, kch thch hucthsacng th hDc tVp v lng tucthnang tin

cuthhoia HS, rn luy,n cho HS n4ng lucthnangc di`n /t sucthnang hiu bit cuthhoia mnh v hiu

di`n /t cuthhoia ngucth%i khc.

T/o mi tructh%ng HS gip \ nhau trong hDc tVp. HS km c i>u ki,n

hDc tVp cc b/n trong nhm, c i>u ki,n tin b trong qu trnh hon

thnh cc nhi,m vuthnang ucth6c giao.

Gip GV thu nhVn tucthsacc th%i nhi>u thng tin phn h0i tucthhuyen pha ngucth%i hDc,

duy tr sucthnang ch cuthhoia HS; gip kim sot hnh vi cuthhoia HS v qun l l"p hDc.

y GV ging nhucth ngucth%i t@ chucthsacc tm ti, cn HS ging nhucth ngucth%i tucthnang lucthnangc

pht hi,n kin thucthsacc m"i, v vVy kt thc cuc m tho/i, HS c ucth6c

ni>m vui cuthhoia sucthnang khm ph, vucthhuyena nHm ucth6c kin thucthsacc m"i, vucthhuyena nHm ucth6c

cch thucthsacc i t"i kin thucthsacc , tructhEng thnh thm mt bucth"c v> trnh tucth

duy. Cui o/n m tho/i, GV cn bit vVn duthnangng cc kin cuthhoia HS kt

luVn v=n > t ra, c b@ sung, chcnh l khi cn thit. Lm ucth6c nhucth vVy,

HS cng hucthsacng th, tucthnang tin v th=y trong kt luVn cuthhoia thy c phn ng

gp kin cuthhoia mnh.

DYn dHt theo phucthQng php v=n p tm ti nhucth trn r rng m=t nhi>u

th%i gian hQn phucthQng php thuyt trnh ging gii nhucthng kin thucthsacc HS

lanh hi ucth6c s| chHc chHn hQn nhi>u.


4. Hn ch

H/n ch l"n nh=t cuthhoia phucthQng php v=n p l r=t kh so/n tho v sucthhoi

duthnangng h, thng cu h8i g6i mE v dYn dHt HS theo mt chuthhoi > nh=t qun.

V vVy i h8i GV phi c sucthnang chu[n b: r=t cng phu, nu khng, kin thucthsacc

m HS thu nhVn ucth6c qua trao @i s| thiu tnh h, thng, tn m/n, thVm

ch vuthnangn vt.

Nu GV chu[n b: h, thng cu h8i khng tt, s| dYn n tnh tr/ng t cu

h8i khng r muthnangc ch, t cu h8i m HS d` dng tr l%i c hoc khng.

Hi,n nay, nhi>u GV thucth%ng gp kh kh4n khi xy ducthnangng h, thng cu h8i

do khng nHm chHc trnh cuthhoia HS, v vVy thucth%ng ngay sau khi t cu

h8i l nu ngay g6i cu tr l%i khin HS rQi vo tr/ng thi b: ng,

khng thucthnangc sucthnang lm vi,c, chc l/i vo g6i cuthhoia GV.

Kh kim sot qu trnh hDc tVp cuthhoia HS (c nhi>u tnh hung b=t ng%

trong cu tr l%i thVm ch cu h8i tucthhuyen pha cuthhoia ngucth%i hDc, gi% hDc d` l,ch

hucth"ng do cu h8i vuthnangn vt, khng nh=t qun).

Kh so/n v xy ducthnangng p n cho cc cu h8i mE (v phucthQng n tr l%i

cuthhoia HS s| khng ging nhau).

5. Mt s lu

Khi so/n cc cu h8i, GV cn lucthu cc yu cu sau y:

Cu h8i phi c ni dung chnh xc, r rng, st v"i muthnangc ch, yu cu cuthhoia

bi hDc, khng lm cho ngucth%i hDc c th hiu theo nhi>u cch khc nhau.

Cu h8i phi st v"i tucthhuyenng lo/i Di tucth-ng HS, nghaa l phi c nhi>u cu

h8i E cc mucthsacc khc nhau, khng qu d` v cuthngang khng qu kh. GV

c kinh nghi,m thucth%ng t8 ra cho HS th=y cc cu h8i >u c tm quan

trDng v kh nhucth nhau ( HS yu c th tr l%i ucth6c nhucthngang cu h8i

vucthhuyena sucthsacc m khng c cm gic tucthnang ti rbng mnh chc c th tr l%i ucth6c

nhucthngang cu h8i d` v khng quan trDng).

Cng mt ni dung hDc tVp, cng mt muthnangc ch nhucth nhau, GV c th sucthhoi

duthnangng nhi>u d/ng cu h8i v"i nhi>u hnh thucthsacc h8i khc nhau.

Bn c/nh nhucthngang cu h8i chnh cn chu[n b: nhucthngang cu h8i phuthnang (trn cQ

sE ducthnang kin cc cu tr l%i cuthhoia HS, trong c th c nhucthngang cu tr l%i sai)

tu tnh hnh thucthnangc t m g6i , dYn dHt tip.

68 | MODULE THCS 18

Nn ch t cc cu h8i mE HS uctha ra nhi>u phucthQng n tr l%i v

pht huy ucth6c tnh tch cucthnangc, sng t/o cuthhoia HS.

Cu h8i ucth6c GV sucthhoi duthnangng v"i nhucthngang muthnangc ch khc nhau, E nhucthngang khu

khc nhau cuthhoia qu trnh d/y hDc nhucthng quan trDng nh=t v cuthngang kh sucthhoi

duthnangng nh=t l E khu nghin cucthsacu ti li,u m"i. Trong khu d/y bi m"i, cu

h8i ucth6c sucthhoi duthnangng trong nhucthngang phucthQng php khc nhau nhucthng quan

trDng nh=t l trong phucthQng php v=n p.

Lo/i cu hSi vIn p ti hin thucth%ng ucth6c sucthhoi duthnangng khi:

+ HS chu[n b: hDc bi.

+ HS ang thucthnangc hnh, luy,n tVp.

+ HS ang n tVp nhucthngang ti li,u hDc.

Lo/i vIn p gii thch, minh ho ucth6c sucthhoi duthnangng trong cc tructh%ng h6p sau:

+ HS c nhucthngang thng tin cQ bn GV mun HS sucthhoi duthnangng cc thng tin =y

trong nhucthngang tnh hung m"i, phucthsacc t/p hQn.

+ HS ang tham gia gii quyt v=n > t ra.

+ HS ang ucth6c cun ht vo cuc tho luVn si n@i v sng t/o.

Lo/i v=n p tm ti d ucth6c sucthhoi duthnangng ring r|, cuthngang c tc duthnangng kch

thch suy ngha tch cucthnangc. V=n p tm ti l phucthQng php ang cn ucth6c

pht trin rng ri. Mun vVy, GV phi u tucth vo vi,c nng cao ch=t

lucth6ng cc cu h8i, gim s cu h8i c yu cu th=p v> mt nhVn thucthsacc (chc

i h8i ti hi,n cc kin thucthsacc sucthnang ki,n), t4ng dn s cu h8i c yu cu cao

v> mt nhVn thucthsacc (i h8i sucthnang thng hiu, phn tch, t@ng h6p, khi qut

ho, h, thng ho, vVn duthnangng kin thucthsacc hDc).

Sucthnang thnh cng cuthhoia phucthQng php g6i mE v=n p phuthnang thuc nhi>u vo

vi,c xy ducthnangng ucth6c h thDng cu hSi g6i mE thch h6p (v phuthnang thuc vo

ngh, thuVt giao tip, ucthsacng xucthhoi v dYn dHt cuthhoia GV).

6. V d

V duthnang minh ho qua mn Ngucthnga vn:

L"p tucthhuyen Hn Vi,t trong cc v4n bn thQ v4n trung /i Vi,t Nam hoc thQ

?ucth%ng c th gy ra nhucthngang trE ng/i nh=t :nh cho HS khi tip nhVn v

cm thuthnang v4n hDc. ?y chnh l cQ hi GV rn luy,n cho HS nhucthngang


ka n4ng suy ngha, tm hiu v4n bn. N4ng lucthnangc sucth ph/m cuthhoia ngucth%i GV ucth6c

th hi,n qua vi,c uctha cu h8i gip HS suy ngha tm ti v cch t@ chucthsacc

cho HS tch cucthnangc gii quyt nhucthngang cu h8i . Qua h, thng cu h8i, HS

s| c ucth6c nhucthngang :nh hucth"ng cQ bn tm hiu, thucthEng thucthsacc, nh gi

tc ph[m v4n hDc theo ng nguyn tHc tip nhVn ngh, thuVt.

Mt s cu h8i v tnh ch=t cuthhoia tucthhuyenng cu trong phn Dc hiu v4n bn

T&ng Phong Nha (Ngucthnga vn 6):

Cu 1: V sao ng Phong Nha ucth6c coi l ?, nh=t k quan? Cu ny l

cu hucth"ng dYn vucthhuyena khm ph bn ch=t cuthhoia v4n bn, vucthhuyena t/o nhucthngang =n

tucth6ng th[m ma ban u v> nhucthngang ni dung phn nh cuthhoia v4n bn.

Cu 2: Bi v4n c th chia thnh hai hay ba o/n? Nu l hai o/n th

cch chia v ni dung cuthnang th cuthhoia tucthhuyenng o/n l g? Nu l ba o/n th cch

chia v ni dung cuthnang th cuthhoia tucthhuyenng o/n l g? ?y l cu h8i gip HS tm

hiu v pht hi,n b cuthnangc (kt c=u) cuthhoia v4n bn v duthnangng ngh, thuVt cuthhoia

nh v4n qua tucthhuyenng phn v4n bn. Trong cu h8i ny, nu thm yu cu

Gii thch v sao l/i chia o/n nhucth vVy? th cu h8i l/i ucth6c nng ln E

mucthsacc cao hQn mucthsacc vVn duthnangng.

Cu 3: Cnh sHc ng Phong Nha ucth6c miu t theo trnh tucthnang no? Trong

ng c nhucthngang b phVn g v p nhucth th no? ?y l cu h8i g6i tm v

khi qut nhucthngang v=n > ni dung v ngh, thuVt cuthhoia v4n bn.

V duthnang minh ho qua mn Ton:

Khi luy,n tVp v> h, thucthsacc v> c/nh v ucth%ng cao trong tam gic vung

(Hnh hDc l"p 9) c th yu cu HS tnh x, y trong hnh v| bn.

Khi hucth"ng dYn HS gii bi ton ny c th sucthhoi duthnangng h, thng cu h8i sau:

Bi ton cho nhucthngang yu t g?

Cn xc :nh yu t no?

Nn tnh /i lucth6ng no tructh"c,

v sao?

Tnh ucth6c y bbng cch no? Sucthhoi

duthnangng h, thucthsacc no?

Tnh ucth6c x bbng cch no? Sucthhoi duthnangng h, thucthsacc no?

C cch no khc tnh x?

70 | MODULE THCS 18

V duthnang minh ho qua mn L%ch sucthhoi:

Hot ng cuthhoia GV Hot ng cuthhoia HS

GV treo bn 0 cc quc

gia c@ /i PhucthQng ?ng.

?i>u khin HS quan

st, tho luVn v> bucthsacc

tranh khHc trn tucth%ng

mt l4ng m E Ai CVp.

Lm bi tVp: ?i>n tip

vo chW trong sQ 0.

Nghin cucthsacu SGK.

Tho luVn nhm.

Quan st tranh v t cu h8i.

Hon thnh sQ 0 sau:

Tn cc

quc gia c

i Phucthng

ng

Thi gian

hnh thnh

!c i"m

v %a bn

Ngh(

chnh

Hot ng 2. Tm tt phucthng php gi m vn p

GV c th tm tHt PPDH ny bbng mt bn 0 tucth duy theo g6i sau:


Hot ng 3. xut mt v d (mt bi dy) v phng php

gi m vn p

GV > xu=t mt v duthnang (mt bi d/y) v> phucthQng php g6i mE v=n p

trong mn hDc cuthhoia mnh.

Hot ng 4. Tho lun nhm phng php gi m vn p v

cc v d xut Hot ng 3

G*i :

VVn duthnangng PPDH ny trong chuyn mn cuthhoia mnh vo cc tnh hung

d/y hDc no: d/y bi m"i, hay luy,n tVp, n tVp, cuthhoing c kin thucthsacc hay

thucthnangc hnh, th nghi,m,?

Nhucthngang kh kh4n khi vVn duthnangng PPDH ny.

V duthnang > xu=t c tructhng cho PPDH ny chuctha hay c th sucthhoi duthnangng v"i PPDH

no khc,

Hot ng 5. nh gi v t nh gi

GV tucthnang rt ra nhucthngang ucthu, nhucth6c im chnh v cch sucthhoi duthnangng phucthQng php

g6i mE v=n p trong mn hDc cuthhoia mnh nhbm /t hi,u qu cao nh=t.

Tham kho bn 0 tucth duy tm tHt PPDH ny i chiu v"i kt qu

Ho/t ng 2 trn.

Cu hi phi c ni

dung chnh xc, r rng,

st vi muthnangc ch

Cng mt ni dung hc t p,

cng mt muthnangc ch nn uctha

cc hnh thucthsacc hi khc nhau

Qu trnh l tucth)ng tc giucthngaa GV

v HS ucth/c thucthnangc hi1n thng

qua h1 th4ng cu hi v cu tr

l5i tucth)ng ucthsacng vi mt chuthhoi 7

nh8t 9nh ucth/c GV :t ra

V8n p ti hi1n, v8n p

gii thch minh ho< v v8n

p tm ti

Bucthc 1: Xc 9nh muthnangc tiu

bi hc v 4i tucth/ng d

72 | MODULE THCS 18

Ni dung 3

TM HIU V PHNG PHP DY HC

PHT HIN V GII QUYT VN

NHIM V

B/n hy Dc ka nhucthngang thng tim cQ bn cuthhoia Ho/t ng 1 lm r:

1. Bn ch=t cuthhoia phucthQng php d/y hDc pht hi,n v gii quyt v=n >; quy

trnh thucthnangc hi,n n.

2. Chc ra nhucthngang ucthu im, nhucthngang h/n ch v nhucthngang im Cn lucthu v>

phucthQng php d/y hDc pht hi,n v gii quyt v=n >.


THNG TIN C BN

Hot ng 1. Tm hiu v phng php dy hc pht hin v

gii quyt vn

Tucthhuyen nhucthngang n4m 1960, GV lm quen v"i thuVt ngucthnga d/y hDc nu v=n >

nhucthng cho n nay vYn chuctha vVn duthnangng thnh th/o. C ngucth%i cho rbng,

thuVt ngucthnga nu v=n > c th gy hiu lm l GV nu ra v=n > HS

gii quyt, do > ngh: thay nu v=n > bbng g6i v=n >. Thucthnangc ra,

tructh"c ht cn tVp ducth6t cho HS kh n4ng pht hi,n v=n > tucthhuyen mt tnh

hung trong hDc tVp hoc trong thucthnangc ti`n. ?y l mt kh n4ng c

nghaa r=t quan trDng i v"i mt con ngucth%i v khng phi d` dng m c

ucth6c. Mt khc, sucthnang thnh /t trong cuc %i khng chc ty thuc vo

n4ng lucthnangc pht hi,n k:p th%i nhucthngang v=n > ny sinh trong thucthnangc ti`n m

bucth"c quan trDng tip theo l gii quyt h6p l nhucthngang v=n > ucth6c t ra.

V vVy, ngy nay ngucth%i ta c xu hucth"ng dng thuVt ngucthnga d/y hDc gii quyt

v=n > hoc d/y hDc nu v gii quyt v=n >, d/y hDc pht hi,n v

gii quyt v=n >.

1. Bn cht ca PPDH pht hin v gii quyt vn

D/y hDc pht hi,n v gii quyt v=n > (PH&GQV?) l PPDH trong

GV t/o ra nhucthngang tnh hung c v=n >, i>u khin HS pht hi,n v=n >,

ho/t ng tucthnang gic, tch cucthnangc, chuthhoi ng, sng t/o gii quyt v=n > v

thng qua chim lanh tri thucthsacc, rn luy,n ka n4ng v /t ucth6c nhucthngang


muthnangc ch hDc tVp khc. ?c tructhng cQ bn cuthhoia d/y hDc PH & GQV? l

tnh hung g6i v=n > v "Tucth duy chc bHt u khi xu=t hi,n tnh hung

c v=n >" (Rubinstein).

Tnh huDng c vIn < (tnh hung g6i v=n >) l mt tnh hung g6i ra

cho HS nhucthngang kh kh4n v> l luVn hay thucthnangc ti`n m hD th=y cn v c kh

n4ng vucth6t qua, nhucthng khng phi ngay tucthsacc khHc bbng mt thuVt gii, m

phi tri qua qu trnh tch cucthnangc suy ngha, ho/t ng bin @i i tucth6ng

ho/t ng hoc i>u chcnh kin thucthsacc syn c.

2. Quy trnh thc hin

Bucthc 1: Pht hin ho.c thm nhp vIn <

Pht hi,n v=n > tucthhuyen mt tnh hung g6i v=n >.

Gii thch v chnh xc ho tnh hung (khi cn thit) hiu ng v=n

> ucth6c t ra.

Pht biu v=n > v t muthnangc tiu gii quyt v=n > .

Bucthc 2: Tm gii php

Tm cch gii quyt v=n > (thucth%ng ucth6c thucthnangc hi,n theo sQ 0 sau):

BHt u

Phn tch v=n >

?> xu=t v thucthnangc hi,n hucth"ng gii quyt

Hnh thnh gii php

Gii php ng

Kt thc

74 | MODULE THCS 18

+ Phn tch vIn xu=t c th ucth6c i>u chcnh khi cn thit. Kt qu cuthhoia

vi,c > xu=t v thucthnangc hi,n hucth"ng gii quyt v=n > l hnh thnh ucth6c

mt gii php.

+ Ki/m tra tnh ng \n cuthhoia gii php: Nu gii php ng th kt thc

ngay, nu khng ng th lp l/i tucthhuyen khu phn tch v=n > cho n khi

tm ucth6c gii php ng. Sau khi tm ra mt gii php, c th tip tuthnangc

tm thm nhucthngang gii php khc, so snh chng v"i nhau tm ra gii

php h6p l nh=t.

Bucthc 3: Trnh by gii php: HS trnh by l/i ton b tucthhuyen vi,c pht biu

v=n > cho t"i gii php. Nu v=n > l mt > bi cho syn th c th

khng cn pht biu l/i v=n >.

Bucthc 4: Nghin cucthsacu su gii php

Tm hiu nhucthngang kh n4ng ucthsacng duthnangng kt qu.

?> xu=t nhucthngang v=n > m"i c lin quan nh% xt tucthQng tucthnang, khi qut ho,

lVt ngucth6c v=n >,... v gii quyt nu c th.

3. u im

PhucthQng php ny gp phn tch cucthnangc vo vi,c rn luy,n tucth duy ph phn,

tucth duy sng t/o cho HS. Trn cQ sE sucthhoi duthnangng vn kin thucthsacc v kinh nghi,m

c, HS s| xem xt, nh gi, th=y ucth6c v=n > cn gii quyt.

?y l phucthQng php pht trin ucth6c kh n4ng tm ti, xem xt ducth"i nhi>u

gc khc nhau. Trong khi PH&GQV?, HS s| huy ng ucth6c tri thucthsacc v

kh n4ng c nhn, kh n4ng h6p tc, trao @i, tho luVn v"i b/n b

tm ra cch gii quyt tt nh=t.

Thng qua vi,c gii quyt v=n >, HS ucth6c lanh hi tri thucthsacc, ka n4ng

v phucthQng php nhVn thucthsacc (gii quyt v=n > khng cn chc thuc


ph/m tr phucthQng php m trE thnh mt muthnangc ch d/y hDc, ucth6c cuthnang

th ho thnh mt muthnangc tiu l pht trin n4ng lucthnangc gii quyt v=n >, mt

n4ng lucthnangc c v: tr hng u con ngucth%i thch ucthsacng ucth6c v"i sucthnang pht trin

cuthhoia x hi).

4. Hn ch

PhucthQng php ny i h8i ngucth%i GV phi u tucth nhi>u th%i gian v cng

sucthsacc, phi c n4ng lucthnangc sucth ph/m tt m"i suy ngha t/o ra ucth6c nhi>u tnh

hung g6i v=n > v hucth"ng dYn HS tm ti PH&GQV?.

Vi,c t@ chucthsacc tit hDc hoc mt phn cuthhoia tit hDc theo phucthQng php

PH&GQV? i h8i phi c nhi>u th%i gian hQn so v"i bnh thucth%ng. HQn

nucthngaa, theo Lecne: Chc c mt s tri thucthsacc v phucthQng php ho/t ng nh=t

:nh, ucth6c lucthnanga chDn kho lo v c cQ sE m"i trE thnh i tucth6ng cuthhoia d/y

hDc nu v=n >.

5. Mt s lu

Lecne khng :nh rbng: S tri thucthsacc v ka n4ng ucth6c HS thu lucth6m trong

qu trnh d/y hDc nu v=n > s| gip hnh thnh nhucthngang c=u trc c bi,t

cuthhoia tucth duy. Nh% nhucthngang tri thucthsacc , t=t c nhucthngang tri thucthsacc khc m HS

lanh hi khng phi tructhnangc tip bbng nhucthngang PPDH nu v=n >, s| ucth6c chuthhoi

th chcnh n l/i, c=u trc l/i. Do , khng nn yu cu HS tucthnang khm

ph t=t c cc tri thucthsacc quy :nh trong chucthQng trnh.

Cho HS PH&GQV? i v"i mt b phVn ni dung hDc tVp, c th c sucthnang

gip \ cuthhoia GV v"i mucthsacc nhi>u t khc nhau. HS ucth6c hDc khng chc

kt qu m i>u quan trDng hQn l c qu trnh PH&GQV?.

HS chcnh n l/i, c=u trc l/i cch nhn i v"i b phVn tri thucthsacc cn l/i

m hD lanh hi khng phi bbng con ucth%ng tucthnang PH&GQV?, thVm ch

c th cuthngang khng phi nghe GV thuyt trnh PH&GQV?. Tc trDng cc

v=n > ngucth%i hDc PH&GQV? so v"i chucthQng trnh tu thuc vo c im

cuthhoia mn hDc, vo i tucth6ng HS v hon cnh cuthnang th. Tuy nhin, phucthQng

hucth"ng chung l: tc trDng phn ni dung ucth6c d/y theo cch HS

PH&GQV? khng chon ht ton b mn hDc nhucthng cuthngang phi uthhoi

ngucth%i hDc bit cch thucthsacc, c ka n4ng gii quyt v=n > v c kh n4ng c=u

trc l/i tri thucthsacc, bit nhn ton b ni dung cn l/i ducth"i d/ng ang trong

qu trnh hnh thnh v pht trin theo cch PH&GQV?.

76 | MODULE THCS 18

GV cn hiu ng cc cch t/o tnh hung g6i v=n > v tVn duthnangng cc cQ

hi t/o ra tnh hung , 0ng th%i t/o i>u ki,n HS tucthnang lucthnangc gii

quyt v=n >. D/y hDc PH&GQV? c th p duthnangng trong cc giai o/n cuthhoia

qu trnh d/y hDc: hnh thnh kin thucthsacc m"i, cuthhoing c kin thucthsacc v ka

n4ng, vVn duthnangng kin thucthsacc. PhucthQng php ny cn hucth"ng t"i mDi i tucth6ng

HS chucthsac khng chc p duthnangng ring cho HS kh gi8i.

Trong d/y hDc PH&GQV? c th phn bi,t 4 mucthsacc :

Mucthsacc 1: GV t v=n >, nu cch gii quyt v=n >. HS thucthnangc hi,n cch gii

quyt v=n > theo sucthnang hucth"ng dYn cuthhoia GV. GV nh gi kt qu lm vi,c

cuthhoia HS.

Mucthsacc 2: GV nu v=n >, g6i HS tm ra cch gii quyt v=n >. HS thucthnangc

hi,n cch gii quyt v=n > v"i sucthnang gip \ cuthhoia GV khi cn. GV v HS cng

nh gi.

Mucthsacc 3: GV cung c=p thng tin t/o tnh hung. HS pht hi,n, nhVn d/ng,

pht biu v=n > ny sinh cn gii quyt, tucthnang lucthnangc > xu=t cc gi thuyt v

lucthnanga chDn cc gii php. HS thucthnangc hi,n k ho/ch gii quyt v=n >. GV v

HS cng nh gi.

Mucthsacc 4: HS tucthnang lucthnangc pht hi,n v=n > ny sinh trong hon cnh cuthhoia mnh

hoc cuthhoia cng 0ng, lucthnanga chDn v=n > phi gii quyt, tucthnang > xu=t ra gi

thuyt, xy ducthnangng k ho/ch gii, thucthnangc hi,n k ho/ch gii, tucthnang nh gi ch=t

lucth6ng v hi,u qu vi,c gii quyt v=n >.

Phn ng GV m"i vVn duthnangng d/y hDc PH&GQV? E mucthsacc 1 v 2. Phi ph=n

=u trong nhi>u tructh%ng h6p c th /t t"i mucthsacc 3 v 4 v lm cho d/y

hDc PH&GQV? trE thnh ph@ bin.

Mt s cch thng duthnangng / to tnh huDng g-i vIn < l: Ducthnang on nh%

nhVn xt tructhnangc quan, thucthnangc hnh hoc ho/t ng thucthnangc ti`n; LVt ngucth6c v=n

>; Xt tucthQng tucthnang; Khi qut ho; Khai thc kin thucthsacc cuthnga, t v=n > dYn

n kin thucthsacc m"i; Gii bi tVp m chuctha bit thuVt gii tructhnangc tip; Tm sai

lm trong l%i gii; Pht hi,n nguyn nhn sai lm v sucthhoia chucthngaa sai lm...

Trong d/y hDc, c r=t nhi>u cQ hi nhucth vVy; do PPDH PH&GQV? c

kh n4ng ucth6c p duthnangng rng ri trong d/y hDc nhbm pht huy tnh chuthhoi

ng, sng t/o cuthhoia HS.


6. V d


V duthnang 1. Dy +%nh l v- t/ng cc gc trong cuthhoia m2t tucthsac gic

Bucthc 1: Pht hi,n hoc thm nhVp v=n >: Mt tam gic b=t k c t@ng

cc gc trong bbng 2v. By gi% cho mt tucthsac gic b=t k, chng h/n ABCD,

li,u ta c th ni g v> t@ng cc gc trong cuthhoia n? Li,u t@ng cc gc trong

cuthhoia n c phi l mt hbng s tucthQng tucthnang nhucth tructh%ng h6p tam gic hay khng?

Bucthc 2: Tm gii php: GV g6i cho HS quy l/ v> quen, uctha vi,c xt tucthsac

gic v> vi,c xt tam gic bbng cch t/o nn nhucthngang tam gic trn hnh v|

tucthQng ucthsacng v"i > bi. Tucthhuyen dYn n vi,c k. ucth%ng cho AC cuthhoia tucthsac gic

ABCD, tucthhuyen HS tm cch gii quyt v=n > t ra.

Bucthc 3: Trnh by gii php: HS trnh by l/i qu trnh gii quyt bi

ton: tucthhuyen vi,c v| hnh, ghi gi thit, kt luVn n vi,c chucthsacng minh.

Bucthc 4: Nghin cucthsacu su gii php: Nghin cucthsacu tructh%ng h6p c bi,t: Tucthsac

gic c 4 gc bbng nhau th mWi gc >u l gc vung.

V duthnang 2. Cch to tnh hu5ng c v6n +-

? thucthnangc hi,n d/y hDc PH&GQV?, im xu=t pht l t/o ra tnh hung c

v=n >. Sau y l mt s cch thng duthnangng t/o ra tnh hung c v=n >.

Cch 1: Ducthnang on nh% nhVn xt tructhnangc quan, nh% thucthnangc hnh hoc ho/t ng

thucthnangc ti`n.

HS quan st (c th ho/t ng o gc, o c/nh, g=p hnh...) mt s cc

tam gic c kch thucth"c, hnh d/ng khc nhau v tm ra c im chung

cuthhoia chng.

Cu tr l%i cuthhoia HS c th l: c ba c/nh, c ba gc,... GV c th t cu

h8i: Tam gic no c t@ng ba gc l"n nh=t trong cc tam gic cho?

Cho HS tucthnang do tho luVn, cng v"i sucthnang dYn dHt cuthhoia GV i n ducthnang on: Cc

tam gic trn c t@ng 3 gc bbng 180

0

.

Cch 2: LVt ngucth6c v=n >.

?t v=n > nghin cucthsacu m,nh > o sau khi chucthsacng minh mt tnh ch=t,

mt :nh l.

78 | MODULE THCS 18

Cch 3: Xem xt tucthQng tucthnang.

Xt nhucthngang php tucthQng tucthnang theo nghaa l chuyn tucthhuyen mt tructh%ng h6p ring

ny sang mt tructh%ng h6p ring khc cuthhoia cng mt ci t@ng qut.

V duthnang: Cho a + b = 2, chucthsacng minh a

2

+ b

2

2

Sau khi chucthsacng minh ucth6c, HS c th nu ln cc bi ton tucthQng tucthnang nhucth:

Cho a + b = 2, tm gi tr: nh8 nh=t cuthhoia a

2

+ b

2

hoc cho a + b + c = 3, chucthsacng minh a

2

+ b

2

+ c

2

3;

Cch 4: Khi qut ho.

V duthnang: Tucthhuyen a

2

b

2

= (a b) (a + b)

a

3

b

3

= (a b)(a

2

+ ab + b

2

)

c th ducthnang on a

n

b

n

= ? (n N; n 2)

Cch 5: Khai thc kin thucthsacc cuthnga t v=n > dYn n kin thucthsacc m"i.

V duthnang minh ho qua mn Ho hc

Nghin cucthsacu th nghi,m: Clo phn ucthsacng v"i dung d:ch ki>m E bi Clo l"p 9.


Nu v=n >: Clo c nhucthngang tnh

ch=t cuthhoia phi kim, ngoi ra clo

cn c tnh ch=t g c bi,t?

Hy nghin cucthsacu th nghi,m clo

tc duthnangng v"i nucth"c v v"i dung

d:ch NaOH.

G6i : Phn ucthsacng ny c g mu

thuYn v"i nhucthngang i>u hDc?

Nhm HS: DYn kh clo vo ng nghi,m

ucthnangng nucth"c c m[u gi=y qu tm v ng

nghi,m ucthnangng dung d:ch NaOH c vi giDt

phenolphtalein.

Quan st hi,n tucth6ng xy ra.

HS nu v=n >: Phn ucthsacng clo v"i dung d:ch

NaOH c mu thuYn v"i tnh ch=t cuthhoia phi

kim hDc khng? hay th nghi,m sai?

HS gii quyt v=n >: Clo c phn ucthsacng v"i

nucth"c t/o thnh 2 axit HCl v HClO. Sau 2

axit ny tip tuthnangc tc duthnangng v"i NaOH t/o



thnh NaCl, NaClO v nucth"c. ?i>u ny l ph

h6p v"i tnh ch=t cuthhoia clo v NaOH hDc.

Kt luVn: Clo phn ucthsacng v"i dung d:ch NaOH

t/o thnh dung d:ch 2 mui.


Trong kh@ cui cuthhoia v4n bn Sang thu, tc gi c nhucthngang cu thQ th

hi,n nhucthngang suy ngYm c nhn. Theo em, l nhucthngang suy ngYm g?

C th ni rbng, nhucthngang tnh hung nhucth trn l tucthQng i tiu biu.

Tuy nhin, khng phi ngay lVp tucthsacc HS c th gii quyt ucth6c tnh

hung v n c lin quan t"i nhi>u mng kin thucthsacc (V4n hDc, Ting Vi,t,

Lm v4n, kin thucthsacc cuc sng...). HS phi bit sucthhoi duthnangng kin thucthsacc cuthnga

c gii quyt tnh hung m"i. GV c th ducthnang kin syn nhucthngang sucthnang hW tr6,

g6i , dYn dHt, nh gi, nhVn xt gip HS gii quyt tnh hung.

Hot ng 2. Tm tt nhng ni dung chnh ca phng php

dy hc pht hin v gii quyt vn


80 | MODULE THCS 18


dy hc pht hin v gii quyt vn

GV > xu=t mt v duthnang (mt bi d/y) v> PPDH pht hi,n v gii quyt v=n

> trong mn hDc m mnh ang ging d/y.

Hot ng 4. Tho lun nhm v phng php dy hc pht hin

v gii quyt vn v cc v d xut Hot ng 3

G*i :






no khc,


GV tucthnang rt ra nhucthngang ucthu, nhucth6c im chnh v cch sucthhoi duthnangng pht hi,n v

gii quyt v=n > trong mn hDc cuthhoia mnh nhbm /t hi,u qu cao nh=t.


Ho/t ng 2 trn.

L PPDH trong GV t


Ni dung 4

TM HIU PHNG PHP DY HC HP TC NHM NH

NHIM V

B/n hy Dc ka thng tin cQ bn cuthhoia Ho/t ng 1 lm r:

1. Bn ch=t cuthhoia phucthQng php d/y hDc h6p tc trong nhm nh8 v quy trnh

thucthnangc hi,n n.

2. Chc ra nhucthngang ucthu im, nhucthngang h/n ch v nhucthngang im cn lucthu v>

phucthQng php d/y hDc h6p tc theo nhm nh8.


THNG TIN C BN

Hot ng 1. Tm hiu v phng php dy hc hp tc trong

nhm nh

N4ng lucthnangc h6p tc ucth6c xem l mt trong nhucthngang n4ng lucthnangc quan trDng cuthhoia

con ngucth%i trong x hi hi,n nay. Chnh v vVy, pht trin n4ng lucthnangc h6p

tc tucthhuyen trong tructh%ng hDc trE thnh mt xu th gio duthnangc trn ton th

gi"i. D/y hDc h6p tc trong nhm nh8 chnh l sucthnang phn nh xu th .

1. Bn cht

PPDH h6p tc trong nhm nh8 cn ucth6c gDi bbng mt s tn khc nhucth

PhucthQng php tho luVn nhm hoc PPDH h6p tc.

?y l mt PPDH m HS ucth6c phn chia thnh tucthhuyenng nhm nh8 ring

bi,t, ch:u trch nhi,m v> mt muthnangc tiu duy nh=t, ucth6c thucthnangc hi,n thng

qua nhi,m vuthnang ring bi,t cuthhoia tucthhuyenng ngucth%i. Cc ho/t ng c nhn ring

bi,t ucth6c t@ chucthsacc l/i, lin kt hucthngau cQ v"i nhau nhbm thucthnangc hi,n mt muthnangc

tiu chung.

PhucthQng php tho luVn nhm ucth6c sucthhoi duthnangng nhbm gip cho mDi HS

tham gia mt cch chuthhoi ng vo qu trnh hDc tVp, t/o cQ hi cho cc em

c th chia s. kin thucthsacc, kinh nghi,m, kin gii quyt cc v=n > c

lin quan n ni dung bi hDc; cQ hi ucth6c giao lucthu, hDc h8i lYn nhau;

cng nhau h6p tc gii quyt nhucthngang nhi,m vuthnang chung.

82 | MODULE THCS 18

2. Quy trnh thc hin

Khi sucthhoi duthnangng PPDH ny, l"p hDc ucth6c chia thnh nhucthngang nhm tucthhuyen 4 n 6

ngucth%i. Ty muthnangc ch sucth ph/m v yu cu cuthhoia v=n > hDc tVp, cc nhm

ucth6c phn chia ngYu nhin hoc c chuthhoi :nh, ucth6c duy tr @n :nh trong

c tit hDc hoc thay @i theo tucthhuyenng ho/t ng, tucthhuyenng phn cuthhoia tit hDc;

cc nhm ucth6c giao cng hoc ucth6c giao nhi,m vuthnang khc nhau.

C=u t/o cuthhoia mt ho/t ng theo nhm (trong mt phn cuthhoia tit hDc,

hoc mt tit, mt bu@i) c th nhucth sau:

Bucthc 1: Lm vi,c chung c l"p

GV gi"i thi,u chuthhoi > tho luVn hoc nu v=n >, xc :nh nhi,m vuthnang

nhVn thucthsacc;

Nu v=n >, xc :nh nhi,m vuthnang nhVn thucthsacc;

T@ chucthsacc cc nhm, giao nhi,m vuthnang cho cc nhm, quy :nh th%i gian v

phn cng v: tr lm vi,c cho cc nhm;

Hucth"ng dYn cch lm vi,c theo nhm (nu cn).

Bucthc 2: Lm vi,c theo nhm

Phn cng trong nhm, tucthhuyenng c nhn lm vi,c c lVp;

Trao @i kin, tho luVn trong nhm;

Cucthhoi /i di,n trnh by kt qu lm vi,c cuthhoia nhm.

Bucthc 3: Tho luVn, t@ng kt tructh"c ton l"p

?/i di,n tucthhuyenng nhm trnh by kt qu tho luVn cuthhoia nhm;

Cc nhm khc quan st, lHng nghe, ch=t v=n, bnh luVn v b@ sung kin;

GV t@ng kt v nhVn xt, t v=n > cho bi tip theo hoc v=n > tip theo.

3. u im

HS ucth6c hDc cch cng tc trn nhi>u phucthQng di,n.

HS ucth6c nu quan im cuthhoia mnh, ucth6c nghe quan im cuthhoia b/n khc

trong nhm, trong l"p; ucth6c trao @i, bn luVn v> cc kin khc nhau v

uctha ra l%i gii ti ucthu cho nhi,m vuthnang ucth6c giao cho nhm. Qua cch hDc ,

kin thucthsacc cuthhoia HS s| b"t phn chuthhoi quan, phin di,n, lm t4ng tnh khch

quan khoa hDc, tucth duy ph phn cuthhoia HS ucth6c rn luy,n v pht trin.


Cc thnh vin trong nhm chia s. cc suy ngha, b4n kho4n, kinh

nghi,m, hiu bit cuthhoia bn thn, cng nhau xy ducthnangng nhVn thucthsacc, thi

m"i v hDc h8i lYn nhau. Kin thucthsacc trE nn su sHc, b>n vucthngang, d` nh" v

nh" nhanh hQn do ucth6c giao lucthu, hDc h8i giucthngaa cc thnh vin trong

nhm, ucth6c tham gia trao @i, trnh by v=n > nu ra. HS ho hucthsacng khi

c sucthnang ng gp cuthhoia mnh vo thnh cng chung cuthhoia c l"p.

Nh% khng kh tho luVn cEi mE nn HS, c bi,t l nhucthngang em nht

nht, trE nn b/o d/n hQn; cc em hDc ucth6c cch trnh by kin cuthhoia

mnh, bit lHng nghe c ph phn kin cuthhoia b/n; tucthhuyen , gip tr. d` ho

nhVp vo cng 0ng nhm, t/o cho cc em sucthnang tucthnang tin, hucthsacng th trong hDc

tVp v sinh ho/t.

Vn hiu bit v kinh nghi,m x hi cuthhoia HS thm phong ph; ka n4ng

giao tip, ka n4ng h6p tc cuthhoia HS ucth6c pht trin.

4. Hn ch

Mt s HS do nht nht hoc v mt s l do no khng tham gia vo

ho/t ng chung cuthhoia nhm. Nu khng phn cng h6p l, chc c mt vi

HS hDc kh tham gia, cn a s HS khc khng ho/t ng.

kin cc nhm c th qu phn tn hoc mu thuYn gay gHt v"i nhau

(nh=t l i v"i cc mn khoa hDc x hi).

Th%i gian c th b: ko di.

V"i nhucthngang l"p c sa s ng hoc l"p hDc chVt hp, bn gh kh di

chuyn th kh t@ chucthsacc ho/t ng nhm. Khi tranh luVn, d ` dYn t"i l"p

0n o, nh hucthEng n cc l"p khc.

5. Mt s lu

C nhi>u cch chia nhm, c th theo s im danh, theo mu sHc, theo

biu tucth6ng, theo gi"i tnh, theo v: tr ng0i hoc c cng sucthnang lucthnanga chDn,

Quy m nhm c th l"n hoc nh8, tu theo nhi,m vuthnang. Tuy nhin, nhm

thucth%ng tucthhuyen 3 5 HS l ph h6p.

Cn quy :nh r th%i gian tho luVn nhm v trnh by kt qu tho luVn

cho cc nhm.

Khi lm vi,c theo nhm, cc nhm c th tucthnang bu ra nhm tructhEng nu

th=y cn. Cc thnh vin trong nhm c th lun phin nhau lm nhm

tructhEng. Nhm tructhEng phn cng cho mWi nhm vin thucthnangc hi,n mt phn

cng vi,c.

84 | MODULE THCS 18

Kt qu tho luVn c th ucth6c trnh by ducth"i nhi>u hnh thucthsacc (bbng l%i,

bbng tranh v|, bbng tiu ph[m, bbng v4n bn vit trn gi=y to,...); c th

do mt ngucth%i thay mt nhm trnh by hoc c th nhi>u ngucth%i trnh

by, mWi ngucth%i mt o/n ni tip nhau.

Trong sut qu trnh HS tho luVn, GV cn n cc nhm, quan st, lHng

nghe, g6i , gip \ HS khi cn thit.

Trong nhm nh8, mWi thnh vin >u ucth6c ho/t ng tch cucthnangc, khng

th l/i vo mt vi ngucth%i n4ng ng v n@i tri hQn. Cc thnh vin

trong nhm gip nhau tm hiu v=n > trong khng kh thi ua v"i cc

nhm khc. Kt qu lm vi,c cuthhoia mWi nhm s| ng gp vo kt qu

chung cuthhoia c l"p. ? trnh by kt qu lm vi,c cuthhoia nhm tructh"c ton l"p,

nhm c th cucthhoi ra mt /i di,n hoc c th phn cng mWi nhm vin

trnh by mt phn nu nhi,m vuthnang ucth6c giao l kh phucthsacc t/p.

Tu theo tucthhuyenng nhi,m vuthnang hDc tVp m sucthhoi duthnangng hnh thucthsacc HS lm vi,c c

nhn hoc ho/t ng nhm cho ph h6p, khng nn thucthnangc hi,n PPDH

ny mt cch hnh thucthsacc. Khng nn l/m duthnangng ho/t ng nhm v cn >

phng xu hucth"ng hnh thucthsacc (trnh li suy ngha: @i m"i PPDH l phi sucthhoi

duthnangng ho/t ng nhm). Chc nhucthngang ho/t ng i h8i sucthnang phi h6p cuthhoia

cc c nhn nhi,m vuthnang hon thnh nhanh chng hQn, hi,u qu hQn

ho/t ng c nhn m"i nn sucthhoi duthnangng phucthQng php ny.

T/o i>u ki,n cc nhm tucthnang nh gi lYn nhau hoc c l"p cng nh gi.

PPDH h6p tc trong nhm nh8 cho php cc thnh vin trong nhm

chia s. cc suy ngha, b4n kho4n, kinh nghi,m, hiu bit cuthhoia bn thn,

cng nhau xy ducthnangng nhVn thucthsacc, thi m"i. Bbng cch ni ra nhucthngang

i>u ang ngha, mWi ngucth%i c th nhVn r trnh hiu bit cuthhoia mnh v>

chuthhoi > nu ra, th=y mnh cn hDc h8i thm nhucthngang g. Bi hDc trE thnh

qu trnh hDc h8i lYn nhau chucthsac khng phi chc l sucthnang tip nhVn thuthnang ng

tucthhuyen GV. Thnh cng cuthhoia l"p hDc phuthnang thuc vo sucthnang nhi,t tnh tham gia cuthhoia

mDi thnh vin, v vVy phucthQng php ny cn ucth6c gDi l phucthQng php

huy ng mDi ngucth%i cng tham gia, hoc rt gDn l phucthQng php cng

tham gia.

Cc cch thnh lp nhm

C r=t nhi>u cch thnh lVp nhm theo cc tiu ch khc nhau, khng

nn p duthnangng mt tiu ch duy nh=t trong c n4m hDc. Bng sau y trnh

by 10 cch theo cc tiu ch khc nhau.


Tiu ch Cch thucthnangc hi0n uchoau, nhucth*c i"m

1. Cc nhm g0m

nhucthngang ngucth%i tucthnang

nguy,n, chung

mi quan tm

?i v"i HS th y l cch d` ch:u nh=t thnh lVp

nhm, m bo cng vi,c thnh cng nhanh nh=t.

D` t/o ra sucthnang tch bi,t giucthngaa cc nhm trong l"p, v vVy cch

t/o nhm nhucth th ny khng nn l kh n4ng duy nh=t.

2. Cc nhm

ngYu nhin

Bbng cch m s, pht th., gHp th4m, sHp xp theo mu sHc...

Cc nhm lun lun m"i s| m bo l t=t c cc HS >u

c th hDc tVp chung nhm v"i t=t c cc HS khc.

Nguy cQ c truthnangc trc s| t4ng cao, HS phi s"m lm quen

v"i vi,c th=y rbng cch lVp nhm nhucth vVy l bnh thucth%ng.

3. Nhm ghp

hnh

X nh8 mt bucthsacc tranh hoc cc t% ti li,u cn xucthhoi l HS ucth6c

pht cc mYu x nh8, nhucthngang HS ghp thnh bucthsacc tranh hoc

t% ti li,u s| t/o thnh nhm.

Cch t/o lVp nhm kiu vui chQi, khng gy ra sucthnang i :ch.

Cn mt t chi ph chu[n b: v cn nhi>u th%i gian hQn

t/o lVp nhm.

4. Cc nhm v"i

nhucthngang c im

chung

V duthnang: T=t c nhucthngang HS cng sinh ra trong ma ng, ma

xun, ma h hoc ma thu s| t/o thnh nhm.

T/o lVp nhm mt cch c o, t/o ra ni>m vui cho HS

c th bit nhau r hQn.

Cch lm ny m=t i tnh c o nu ucth6c sucthhoi duthnangng

thucth%ng xuyn.

5. Cc nhm c

:nh trong mt

th%i gian di

Cc nhm ucth6c duy tr trong mt s tun hoc mt s

thng, cc nhm ny thVm ch c th ucth6c t tn ring.

Cch lm ny ucth6c chucthsacng t8 hi,u qu tt trong nhucthngang

nhm hDc tVp c nhi>u v=n >.

Sau khi quen nhau mt th%i gian di th vi,c lVp cc

nhm m"i s| kh kh4n.

6. Nhm c HS kh

hW tr6 HS yu

Nhucthngang HS kh gi8i trong l"p cng luy,n tVp v"i cc HS yu

hQn v m nhVn nhi,m vuthnang cuthhoia ngucth%i hucth"ng dYn.

86 | MODULE THCS 18

Tiu ch Cch thucthnangc hi0n uchoau, nhucth*c i"m

T=t c >u ucth6c l6i. Nhucthngang HS gi8i m nhVn trch

nhi,m, nhucthngang HS yu ucth6c gip \.

Ngoi vi,c m=t nhi>u th%i gian th chc c t nhucth6c im,

tructhhuyen phi nhucthngang HS gi8i hucth"ng dYn sai.

7. Phn chia theo

n4ng lucthnangc hDc tVp

khc nhau

Nhucthngang HS yu hQn s| xucthhoi l cc bi tVp cQ bn nhucthngang HS c

bi,t gi8i s| nhVn ucth6c thm nhucthngang bi tVp b@ sung.

HS c th tucthnang xc :nh muthnangc ch cuthhoia mnh. V duthnang ai b: im

km trong mn Ton th c th tVp trung vo mt s t bi tVp.

Cch lm ny dYn n kt qu l nhm hDc tVp cm th=y

b: chia thnh nhucthngang HS thng minh v nhucthngang HS km.

8. Phn chia theo

cc d/ng hDc tVp

?ucth6c p duthnangng thucth%ng xuyn khi hDc tVp theo tnh hung;

nhucthngang HS thch hDc tVp v"i hnh nh, m thanh hoc biu

tucth6ng s| nhVn ucth6c nhucthngang bi tVp tucthQng ucthsacng.

HS s| bit cc em thuc d/ng hDc tVp nhucth th no.

HS chc hDc nhucthngang g mnh thch v b8 qua nhucthngang ni dung khc.

9. Nhm v"i cc

bi tVp khc

nhau

V duthnang, trong khun kh@ mt ducthnang n, mt s HS s| kho st mt

x nghi,p, mt s khc kho st mt cQ sE ch4m sc x hi

T/o i>u ki,n hDc tVp theo kinh nghi,m i v"i nhucthngang g

c bi,t quan tm.

Thucth%ng chc c th p duthnangng trong khun kh@ mt ducthnang n l"n.

10. Phn chia HS

nam v nucthnga

C th thch h6p nu hDc v> nhucthngang chuthhoi > c tructhng cho

HS nam v nucthnga, v duthnang trong ging d/y v> tnh duthnangc, chuthhoi > lucthnanga

chDn ngh> nghi,p.

Nu b: l/m duthnangng s| dYn n m=t bnh ng nam nucthnga.

6. V d


Khi d/y bi uchoa"c v bi E l"p 6, sau khi hDc xong :nh nghaa v cch tm

ucth"c v bi cuthhoia mt s, cuthhoing c, GV c th thucthnangc hi,n ho/t ng nhm:

Chia l"p thnh cc nhm tucthhuyen 3 n 4 HS. Cc nhm c s thucthsac tucthnang l. gii bi


E phiu s 1, nhm c s thucthsac tucthnang chyn gii bi E phiu s 2. Th%i gian lm

vi,c nhm l 2 pht.

Phi3u s 1. Cho cc s: 1; 12; 14; 2; 18; 23; 0; 3.

a) Vit tVp h6p A cc s thuc dy trn l bi cuthhoia 6.

b) Vit tVp h6p B cc s thuc dy trn l ucth"c cuthhoia 6.

Phi3u s 2. Cho mn = 30 v x = 7t (m, n, x, t N*).

Hy i>n vo chW trng cc tucthhuyen "ucth"c", "bi" ucth6c cc kt luVn ng.

a/ m l ............... cuthhoia 30. b/ 30 l ............... cuthhoia m.

c/ x l ................. cuthhoia t. d/ x l ................. cuthhoia 7t.

e/ t l ................. cuthhoia x. g/ 7 l ................. cuthhoia x.

Sau khi thucthnangc hi,n xong ho/t ng trn, GV c th t@ chucthsacc tr chQi: Thi

nhm no nhanh hn bbng cch chia l"p thnh cc nhm, mWi nhm 4

HS gii bi: Tm cc bi cuthhoia 9 l"n hQn 20 v nh8 hQn 200.

Sau khong 2 pht, gDi /i di,n ba nhm c kt qu nhanh nh=t ln ghi

kt qu ln bng. Nhm no ghi ucth6c nhi>u kt qu ng nh=t, nhm

s| thHng.

V duthnang minh ho qua mn Gio duthnangc Cng dn:

Khi d/y bi 14: Bo v, mi tructh%ng v ti nguyn thin nhin (Gio duthnangc

Cng dn l"p 7), sau khi cho HS quan st cc bucthsacc nh hoc b4ng hnh v>

cnh luthnga luthnangt, h/n hn, chy ructhhuyenng, nhi`m khng kh,... GV c th t@ chucthsacc

cho HS tho luVn nhm theo cc cu h8i sau:

+ Em ngha g khi xem cc cnh trn?

+ Luthnga luthnangt, h/n hn, chy ructhhuyenng, nhi` m khng kh,... nh hucthEng n cuc

sng cuthhoia con ngucth%i nhucth th no?

+ Nguyn nhn no dYn n nhucthngang thm ho/ ?

+ Chng ta cn lm g h/n ch, ng4n ngucthhuyena cc thm ho/ ?

V duthnang minh ho qua mn Ho hc:

V duthnang 1. Nhm HS nghin cucthsacu tnh ch=t chung cuthhoia axit (axit tc duthnangng v"i

bazQ) thng qua th nghi,m nghin cucthsacu dung d:ch H

2

SO

4

tc duthnangng v"i

Cu(OH)

2

v NaOH.

88 | MODULE THCS 18

Ho/t ng cuthhoia HS c th l:

Cc thnh vin Nhi0m vuthnang

Nhm tructhEng Phn cng, i>u khin

Thucth k Ghi chp kt qu bo co cuthhoia cc thnh vin

Cc thnh vin

Quan st tr/ng thi, mu sHc cuthhoia dung d:ch

H

2

SO

4

, Cu(OH)

2

, NaOH rHn

Thnh vin 1

TN1: Nh8 tucthhuyen tucthhuyen dung d:ch H

2

SO

4

vo ng nghi,m

ucthnangng Cu(OH)

2

Thnh vin 2

TN2: Nh8 tucthhuyen tucthhuyen dung d:ch H

2

SO

4

vo ng nghi,m

ucthnangng NaOH

Cc thnh vin

Quan st, m t hi,n tucth6ng xy ra E TN1 v TN2.

Gii thch v rt ra kt luVn

Nhm tructhEng

Chc /o tho luVn. Rt ra kt luVn chung. Bo co

kt qu cuthhoia nhm

V duthnang 2. T@ chucthsacc ho/t ng nhm trong bi thucthnangc hnh Tnh ch=t cuthhoia axit

axetic v ructh6u etylic, th nghi,m 2.

Ho/t ng cuthhoia GV v nhm HS:

Hot ng cuthhoia GV Hot ng cuthhoia nhm HS

Yu cu HS bo co ni dung

chu[n b: tructh"c E nh.

GV hon thi,n v cht l/i trn bng

phuthnang (bn trong hoc mn hnh).

?/i di,n nhm HS bo co kt qu chu[n

b: E nh.

Nu muthnangc ch cuthhoia th nghi,m: Thucthnangc hnh

v> tnh ch=t cuthhoia C

2

H

5

OH v CH

3

COOH.

Duthnangng cuthnang, ho ch=t cn thit. Cch lHp

duthnangng cuthnang.

Cch tin hnh v mt s ka thuVt cn

ch . V duthnang cch lHp nt cao su c ng

dYn xuyn qua, cch un hWn h6p phn


Hot ng cuthhoia GV Hot ng cuthhoia nhm HS

Yu cu HS quan st tr/ng thi cuthhoia

cc dung d:ch H

2

SO

4

, C

2

H

5

OH v

CH

3

COOH.

GV i t"i cc nhm quan st v hW

tr6 nu cn.

ucthsacng, cch thu sn ph[m v lm r sn

ph[m, th%i gian tin hnh th nghi,m...

HS lHng nghe, gp b@ sung.

1 2 HS Dc l/i ni dung tructh"c khi tin

hnh th nghi,m.

Nhm HS tin hnh th nghi,m.

Nhm tructhEng phn cng cho cc nhm

vin cc nhi,m vuthnang:

Quan st tr/ng thi, mu sHc cuthhoia cc

dung d:ch H

2

SO

4

, C

2

H

5

OH v CH

3

COOH.

LHp duthnangng cuthnang nhucth hnh v|.

L=y ho ch=t theo :nh lucth6ng ghi.

Chm n c0n.

?t nng ng nghi,m.

Quan st hi,n tucth6ng phn ucthsacng: Ch

phn ch=t l8ng thu ucth6c E ng nghi,m

t trong cc nucth"c l/nh.

Kt thc qu trnh un, l=y ng nghi,m

ucthnangng sn ph[m v thm vo 2ml nucth"c

mui bo ho r0i lHc nh.

Quan st l"p ch=t l8ng E pha trn.

HS ghi l/i hi,n tucth6ng, gii thch v vit

PTHH (do c nhm tho luVn).



V sao qun Nguyn mc d th=t

b/i nng n> trong cuc chin tranh

xm lucth6c ?/i Vi,t ln thucthsac hai l/i quyt

tm xm lucth6c ?/i Vi,t ln thucthsac ba?

Nghin cucthsacu SGK.

Tho luVn nhm.

Tr l%i:

90 | MODULE THCS 18


+ Chuctha tucthhuyen b8 0 bnh tructh"ng xung

phucthQng Nam.

+ Qun Nguyn mun tr th, ructhhoia nhuthnangc


Trong v4n bn Vucth-t thc E l"p 6 c ba o/n tucthQng i c lVp, GV c

th giao nhi,m vuthnang cho tucthhuyenng nhm Dc hiu v trnh by nhucthngang nh

gi, nhVn xt cuthhoia mnh v> i tucth6ng miu t, im nhn trn thuVt v

ngh, thuVt miu t trong tucthhuyenng o/n. Sau cc nhm c th nhVn xt,

nh gi chnh xc trong cch hiu, cch di`n /t cuthhoia nhau. Cui

cng, GV tVp h6p t@ng kt l/i cc kin v nh gi chnh xc cuthhoia

cc cu tr l%i.

Trong gi% hDc v> mt v4n bn nhVt duthnangng, c th nu v=n > cho HS tho

luVn nghaa cuthhoia v=n > th%i sucthnang m v4n bn uctha ra v cch ucthsacng xucthhoi cn

thit cuthhoia c nhn tructh"c v=n > .


dy hc hp tc trong nhm nh




dy hc hp tc nhm nh

GV > xu=t mt v duthnang (mt bi d/y) v> PPDH h6p tc nhm nh8 trong

mn hDc m mnh ang ging d/y.

Hot ng 4. Tho lun nhm v phng php dy hc hp tc

nhm nh v cc v d xut Hot ng 3

G*i :






no khc,


GV tucthnang rt ra nhucthngang ucthu, nhucth6c im chnh v cch sucthhoi duthnangng PPDH h6p tc

nhm nh8 trong mn hDc cuthhoia mnh nhbm /t hi,u qu cao nh=t.


Ho/t ng 2 trn.

HS ucth/c phn chia thnh tucthhuyenng nhm nh

ring bi1t, ch9u trch nhi1n v7 mt muthnangc tiu

duy nh8t, ucth/c thucthnangc hi1n thng qua nhi1m

vuthnang ring bi1t cuthhoia tucthhuyenng ngucth5i

PPDH h/p tc trong nhm nh cn ucth/c

gi l: Phucth)ng php tho lu n nhm ho:c

phucth)ng php d

92 | MODULE THCS 18

Ni dung 5

TM HIU V PHNG PHP DY HC TRC QUAN

Hot ng 1. c, tm hiu v phng php dy hc trc quan

NHIM V

B/n hy Dc ka thng tin cQ bn cuthhoia Ho/t ng 1 lm r:

1. Bn ch=t cuthhoia phucthQng php d/y hDc tructhnangc quan v quy trnh thucthnangc hi,n n.

2. Chc ra nhucthngang ucthu im, nhucthngang h/n ch v im cn lucthu v> phucthQng php

d/y hDc tructhnangc quan.


THNG TIN C BN

1. Bn cht ca phng php dy hc trc quan

D/y hDc tructhnangc quan (hay cn gDi l trnh by tructhnangc quan) l phucthQng php

sucthhoi duthnangng nhucthngang phucthQng ti,n tructhnangc quan, phucthQng ti,n ka thuVt d/y hDc

tructh"c, trong v sau khi nHm ti li,u m"i, khi n tVp, cuthhoing c, h, thng

ho v kim tra tri thucthsacc, ka n4ng, ka xo.

PPDH tructhnangc quan ucth6c th hi,n ducth"i hai hnh thucthsacc l minh ho/ v trnh by:

Minh ho/ thucth%ng tructhng by nhucthngang 0 dng tructhnangc quan c tnh ch=t minh

ho/ nhucth bn mYu, bn 0, bucthsacc tranh, tranh chn dung, hnh v| trn bng...

Trnh by thucth%ng gHn li>n v"i vi,c trnh by th nghi,m, nhucthngang thit b: ka

thuVt, chiu phim n chiu, phim i,n nh, b4ng video. Trnh by th

nghi,m l trnh by m hnh /i di,n cho hi,n thucthnangc khch quan ucth6c lucthnanga

chDn c[n thVn v> mt sucth ph/m. N l cQ sE, l im xu=t pht cho qu

trnh nhVn thucthsacc hDc tVp cuthhoia HS, l cu ni giucthngaa l thuyt v thucthnangc ti`n.

Thng qua sucthnang trnh by cuthhoia GV m HS khng chc lanh hi d` dng tri thucthsacc

m cn gip hD hDc tVp ucth6c nhucthngang thao tc mYu cuthhoia GV, tucthhuyen hnh

thnh ka n4ng, ka xo

2. Quy trnh thc hin phng php dy hc trc quan

GV treo nhucthngang 0 dng tructhnangc quan c tnh ch=t minh ho/ hoc gi"i thi,u

v> cc vVt duthnangng th nghi,m, cc thit b: ka thuVt Nu yu cu :nh

hucth"ng cho sucthnang quan st cuthhoia HS.


GV trnh by cc ni dung trong lucth6c 0, sQ 0, bn 0, tin hnh lm th

nghi,m, trnh chiu cc thit b: ka thuVt, phim n chiu, phim i,n nh

GV yu cu mt hoc mt s HS trnh by l/i, gii thch ni dung sQ 0,

biu 0, trnh by nhucthngang g thu nhVn ucth6c qua th nghi,m hoc qua

nhucthngang phucthQng ti,n ka thuVt, phim n chiu, phim i,n nh.

Tucthhuyen nhucthngang chi tit, thng tin HS thu ucth6c tucthhuyen phucthQng ti,n tructhnangc quan, GV nu

cu h8i yu cu HS rt ra kt luVn khi qut v> v=n > m phucthQng ti,n

tructhnangc quan cn chuyn ti.

3. u im ca phng php dy hc trc quan

Nguyn tHc tructhnangc quan l mt trong nhucthngang nguyn tHc cQ bn cuthhoia l luVn

d/y hDc nhbm t/o cho HS nhucthngang biu tucth6ng v hnh thnh cc khi ni,m

trn cQ sE tructhnangc tip quan st hi,n vVt ang hDc hay 0 dng tructhnangc quan

minh ho/ sucthnang vVt. ?0 dng tructhnangc quan l chW ducthnanga hiu su sHc bn ch=t

kin thucthsacc, l phucthQng ti,n c hi,u lucthnangc hnh thnh cc khi ni,m, gip HS

nHm vucthngang cc quy luVt cuthhoia sucthnang pht trin x hi. V duthnang, khi nghin cucthsacu bucthsacc

tranh: Hnh v| trn vch hang, HS khng chc c biu tucth6ng v> s4n bHn

l cng vi,c thucth%ng xuyn v hng u cuthhoia th: tc, m cn hiu: nh% ch

t/o cung tn, con ngucth%i chuyn hn tucthhuyen hnh thucthsacc sn b\t sang sn b\n,

c hi,u qu kinh t cao hQn. ?i>u gip HS bit sucthnang thay @i trong %i

sng vVt ch=t cuthhoia con ngucth%i th%i nguyn thu lun gHn cht v"i tin b

trong ka thuVt ch tc cng cuthnang cuthhoia hD.

?0 dng tructhnangc quan c vai tr r=t l"n trong vi,c gip HS nh" ka, hiu su

nhucthngang hnh nh, nhucthngang kin thucthsacc l:ch sucthhoi. Hnh nh ucth6c giucthnga l/i c bi,t

vucthngang chHc trong tr nh" l hnh nh chng ta thu nhVn ucth6c bbng tructhnangc

quan. V vVy, cng v"i vi,c gp phn t/o biu tucth6ng v hnh thnh khi

ni,m l:ch sucthhoi, 0 dng tructhnangc quan cn pht trin kh n4ng quan st, tr

tucthEng tucth6ng, tucth duy v ngn ngucthnga cuthhoia HS.

4. Nhc im ca phng php dy hc trc quan

PhucthQng php ny i h8i nhi>u th%i gian, GV cn tnh ton ka ph h6p

v"i th%i lucth6ng quy :nh.

Nu sucthhoi duthnangng 0 dng tructhnangc quan khng kho s| lm phn tn ch cuthhoia HS,

lm HS khng lanh hi ucth6c nhucthngang ni dung chnh cuthhoia bi hDc.

94 | MODULE THCS 18

Khi sucthhoi duthnangng 0 dng tructhnangc quan, c bi,t l khi quan st tranh nh, cc phim

i,n nh, phim video, nu GV khng :nh hucth"ng cho HS quan st s| d` dYn

n tnh tr/ng HS sa vo nhucthngang chi tit nh8 l., khng quan trDng.

5. Mt s lu khi s dng dng trc quan dy hc

Khi sucthhoi duthnangng 0 dng tructhnangc quan trong d/y hDc, cn ch cc nguyn tHc sau:

Phi c4n cucthsac vo ni dung, yu cu gio duthnangc cuthhoia bi hDc lucthnanga chDn 0

dng tructhnangc quan tucthQng ucthsacng thch h6p. V vVy, cn xy ducthnangng mt h, thng

0 dng tructhnangc quan phong ph, ph h6p v"i tucthhuyenng lo/i bi hDc.

C phucthQng php thch h6p i v"i vi,c sucthhoi duthnangng mWi lo/i 0 dng tructhnangc quan.

Phi m bo ucth6c sucthnang quan st y uthhoi 0 dng tructhnangc quan cuthhoia HS.

Pht huy tnh tch cucthnangc cuthhoia HS khi sucthhoi duthnangng 0 dng tructhnangc quan.

?m bo kt h6p l%i ni v"i vi,c trnh by cc 0 dng tructhnangc quan, 0ng

th%i rn luy,n kh n4ng thucthnangc hnh cuthhoia HS khi xy ducthnangng v sucthhoi duthnangng 0

dng tructhnangc quan (Hp sa bn, v| bn 0, tucth%ng thuVt trn bn 0, miu t

hi,n vVt).

Tu theo yu cu cuthhoia bi hDc v lo/i hnh 0 dng tructhnangc quan m c cc cch

sucthhoi duthnangng khc nhau. Lo/i 0 dng tructhnangc quan treo tucth%ng ucth6c sucthhoi duthnangng

nhi>u nh=t trong d/y hDc hi,n nay l vVt mYu, bn 0, sQ 0, 0 th:, bng

nin biu Tructh"c khi sucthhoi duthnangng chng cn chu[n b: thVt ka (nHm chHc ni

dung, nghaa cuthhoia tucthhuyenng lo/i phuthnangc vuthnang cho ni dung no cuthhoia gi% hDc).

Trong khi ging, cn xc :nh ng th%i im uctha 0 dng tructhnangc quan.

Lo/i 0 dng tructhnangc quan c\ nh8 ucth6c sucthhoi duthnangng ring cho tucthhuyenng HS trong gi%

hDc, trong vi,c tucthnang hDc E nh, GV phi hucth"ng dYn HS sucthhoi duthnangng tt lo/i 0

dng tructhnangc quan ny: quan st ka, tm hiu su sHc ni dung, hon thnh

cc bi tVp, tVp v| bn 0, chucthsac khng phi can theo sch.

Trong d/y hDc mt s mn nhucth L:ch sucthhoi, ?:a l, Sinh hDc, m nh/c, Cng

ngh,, Ma thuVt... E tructh%ng ph@ thng, vi,c kt h6p cht ch| giucthngaa l%i ni

sinh ng v"i sucthhoi duthnangng 0 dng tructhnangc quan l mt trong nhucthngang i>u quan

trDng nh=t thucthnangc hi,n nhi,m vuthnang gio ducth\ng, gio duthnangc v pht trin.

Sucthhoi duthnangng cc 0 dng tructhnangc quan cn theo mt quy trnh h6p l c th

khai thc ti a kin thucthsacc tucthhuyen cc 0 dng tructhnangc quan. Cn chu[n b: cu

h8i/h, thng cu h8i dYn dHt HS quan st v tucthnang khai thc kin thucthsacc.


6. V d minh ho



GV treo lucth6c 0 chin thHng B/ch

?bng n4m 1288. H8i HS: Ducthnanga trn cQ

sE no m Trn Hucthng ?/o xc :nh k

ho/ch phn cng?

GV trnh by bbng bn 0 di`n bin

trVn B/ch ?bng:

Gic s| rt theo hai ucth%ng thu, b:

qun b: ucth%ng L/ng SQn; qun thu:

ucth%ng sng B/ch ?bng.

VVy Trn Hucthng ?/o c k ho/ch g

trong ln phn cng ny?

Chin thHng on thuy>n lucthQng;

Gic lm vo tnh th lng tng;

TrVn Vn ?0n thHng l6i.

HS xem o/n b4ng v> con sng B/ch

?bng.

Trn Hucthng ?/o chDn v chu[n b:

trVn :a E sng B/ch ?bng v:

+ ThHng qun Nam Hn do Ng

Quy>n chc huy n4m 938.

+ ?:a th him trE.

+ Mucthsacc nucth"c ln xung r r,t

V duthnang minh ho qua mn Sinh hc:

Quan st mt s thn bin d/ng

Muthnangc tiu: Quan st ucth6c hnh d/ng v bucth"c u phn nhm cc lo/i

thn bin d/ng, th=y ucth6c chucthsacc n4ng i v"i cy.

Tin hnh:


a. Quan st cc loi cuthhoi, tm .c i/m

chucthsacng tS chng l thn

GV yu cu HS cho bit thn cy c c

im g (c ch0i ngDn, ch0i nch v l)

r0i yu cu HS quan st cc lo/i cuthhoi xem

chng c c im g chucthsacng t8 l thn.

HS t mYu ln bn quan st tm xem

c ch0i, l khng.

HS quan st + tranh nh v g6i cuthhoia

GV chia cuthhoi thnh nhi>u nhm.

96 | MODULE THCS 18


GV lucthu tm cuthhoi su ho c ch0i nch

v gucthhuyenng c ch0i HS quan st thm.

GV cho HS phn chia cc lo/i cuthhoi

thnh nhm ducthnanga trn v: tr cuthhoia n so

v"i mt =t v hnh dng cuthhoi, chucthsacc n4ng.

GV yu cu HS tm nhucthngang c im

ging v khc nhau giucthngaa cc lo/i cuthhoi ny.

GV lucthu HS bc v8 cuthhoia cuthhoi dong

tm dDc cuthhoi c nhucthngang mHt 8 l ch0i

nch, cn cc v8 (hnh v[y) l.

GV cho HS trnh by v tucthnang b@ sung

cho nhau.

GV nhVn xt v t@ng kt: Mt s lo/i

thn bin d/ng lm chucthsacc n4ng khc l

ducthnang tructhnga ch=t khi ra hoa, kt qu.

b. Quan st thn cy xucthng r1ng

GV cho HS quan st thn cy xucthQng

r0ng, tho luVn theo cu h8i:

+ Thn xucthQng r0ng chucthsaca nhi>u nucth"c

c tc duthnangng g?

+ Sng trong i>u ki,n no l bin

thnh gai?

+ Cy xucthQng r0ng thucth%ng sng E u?

+ K tn mt s cy mDng nucth"c.

GV cho HS nghin cucthsacu SGK r0i rt ra

kt luVn.

HS pht hi,n cc c im:

+ ?c im ging nhau: c ch0i, l

l thn.

?>u phnh to chucthsaca ch=t ducthnang tructhnga.

+ ?c im khc nhau: d/ng r`

Cuthhoi gucthhuyenng, dong (c hnh r`) ducth"i mt

=t thn cuthhoi.

?/i di,n nhm ln trnh by kt qu

cuthhoia nhm v nhm khc nhVn xt,

b@ sung.

HS Dc tr.58 SGK. Trao @i nhm

theo 4 cu h8i SGK.

?/i di,n nhm trnh by kt qu,

nhm khc b@ sung.

HS quan st thn, gai, ch0i ngDn cuthhoia

cy xucthQng r0ng. Dng que nhDn chDc

vo thn quan st hi,n tucth6ng

tho luVn nhm.

?/i di,n nhm trnh by kt qu

nhm khc b@ sung.

HS Dc tr.58 SGK sucthhoia chucthngaa kt qu.

Kt luVn: Thn bin d/ng chucthsaca ch=t

ducthnang tructhnga hay ducthnang tructhnga nucth"c cho cy.



dy hc trc quan



dy hc trc quan

GV > xu=t mt v duthnang (mt bi d/y) v> PPDH tructhnangc quan trong mn hDc

m mnh ang ging d/y.

Hot ng 4. Tho lun nhm v phng php dy hc trc

quan v cc v d xut Hot ng 3

G*i :






no khc,

98 | MODULE THCS 18


GV tucthnang rt ra nhucthngang ucthu, nhucth6c im chnh v cch sucthhoi duthnangng PPDH tructhnangc

quan trong mn hDc cuthhoia mnh nhbm /t hi,u qu cao nh=t.


ho/t ng 5.2 trn.

Ni dung 6

TM HIU V PHNG PHP DY HC LUYN TP V THC HNH

Hot ng 1. Tm hiu v phng php dy hc luyn tp v

thc hnh

NHIM V

B/n hy Dc nhucthngang thng tin phn h0i cuthhoia Ho/t ng 1 lm r:

1. Bn ch=t cuthhoia phucthQng php d/y hDc luy,n tVp v thucthnangc hnh; quy trnh

thucthnangc hi,n n.

2. Chc ra nhucthngang ucthu im, nhucthngang h/n ch v nhucthngang im cn lucthu v> phucthQng

php luy,n tVp v thucthnangc hnh.



THNG TIN C BN

1. Bn cht

Luy,n tVp v thucthnangc hnh nhbm cuthhoing c, b@ sung, lm vucthngang chHc thm cc

kin thucthsacc l thuyt. Trong luy,n tVp, ngucth%i ta nh=n m/nh t"i vi,c lp l/i

v"i muthnangc ch hDc thuc nhucthngang o/n thng tin: o/n v4n, thQ, bi ht,

k hi,u, quy tHc, :nh l, cng thucthsacc,... hDc v lm cho vi,c sucthhoi duthnangng ka

n4ng ucth6c thucthnangc hi,n mt cch tucthnang ng, thnh thuthnangc. Trong thucthnangc hnh,

ngucth%i ta khng chc nh=n m/nh vo vi,c hDc thuc m cn nhbm p duthnangng

hay sucthhoi duthnangng mt cch thng minh cc tri thucthsacc thucthnangc hi,n cc nhi,m vuthnang

khc nhau. V th, trong d/y hDc, bn c/nh vi,c cho HS luy,n tVp mt s

chi tit cuthnang th, GV cuthngang cn lucthu cho HS thucthnangc hnh pht trin cc ka n4ng.

2. Quy trnh thc hin

Bucthc 1: Xc :nh ti li,u cho luy,n tVp v thucthnangc hnh

Bucth"c ny bao g0m vi,c tVp trung ch cuthhoia HS v> mt ka n4ng cuthnang th

hoc nhucthngang sucthnang ki,n cn luy,n tVp hoc thucthnangc hnh.

Bucthc 2: Gi"i thi,u m hnh luy,n tVp hoc thucthnangc hnh

Khun mYu HS bHt chucth"c hoc lm theo ucth6c GV gi"i thi,u, c th

thng qua v duthnang cuthnang th.

Bucthc 3: Thucthnangc hnh hoc luy,n tVp sQ b

HS tm hiu v> ti li,u luy,n tVp hoc thucthnangc hnh. HS c th tucthnang thucthhoi ka

n4ng cuthhoia mnh v c th t cu h8i v> nhucthngang ka n4ng . Vi,c nhHc l/i

sQ b c th ucth6c tin hnh trong ho/t ng cuthhoia c l"p v"i sucthnang hucth"ng dYn

cuthhoia GV. Nu luy,n tVp hay thucthnangc hnh mt ka n4ng tucthnang ng th mWi bucth"c

cn c l%i chc dYn cuthnang th. Bi tVp lo/i ny cn ucth6c tip tuthnangc cho t"i khi HS

bit chnh xc hD phi lm g v nhVn r mucthsacc hon thnh m cc em

cn /t ucth6c.

Bucthc 4: Thucthnangc hnh a d/ng

GV uctha ra cc bi tVp i h8i HS phi sucthhoi duthnangng nhi>u kin thucthsacc, :nh l,

cng thucthsacc... Cc bi tVp cng a d/ng th HS cng c cQ hi rn luy,n ka

n4ng, vVn duthnangng cc kin thucthsacc khc nhau gii quyt nhi,m vuthnang t ra.

100 | MODULE THCS 18

Bucthc 5: Bi tVp c nhn

HS c th luy,n tVp, thucthnangc hnh nhucthngang bi tVp c trong SGK hoc sch bi

tVp hoc cc bi tVp tham kho khc nhbm pht trin ka n4ng gii quyt

v=n > v rn luy,n tucth duy.

3. u im

?y l phucthQng php c hi,u qu mE rng sucthnang lin tucthEng v pht trin

cc ka n4ng.

Luy,n tVp v thucthnangc hnh c hi,u qu trong vi,c cuthhoing c tr nh", tinh lDc

v trau chut cc ka n4ng hDc, t/o cQ sE cho vi,c xy ducthnangng ka n4ng

nhVn thucthsacc E mucthsacc cao hQn.

?y l phucthQng php d` thucthnangc hi,n v ucth6c thucthnangc hi,n trong hu ht cc

gi% hDc nhucth mn Ton, Th duthnangc, m nh/c,...

4. Hn ch

Luy,n tVp v thucthnangc hnh c xu hucth"ng lm cho HS nhm chn nu GV

khng nu muthnangc ch mt cch r rng v c sucthnang khuyn khch cao. D` t/o

tm l phuthnang thuc vo mYu, h/n ch sucthnang sng t/o.

Do bn ch=t cuthhoia vi,c nhHc i nhHc l/i nn HS kh c th /t ucth6c sucthnang lanh

l6i v tVp trung, d` t/o nn sucthnang hDc vt, c bi,t l khi chuctha xy ducthnangng ucth6c

sucthnang hiu bit ban u y uthhoi.

5. Mt s lu

Luy,n tVp v thucthnangc hnh cn phi ucth6c tin hnh thucth%ng xuyn trong

(mt s) p lucthnangc. Cc bi tVp luy,n tVp ucth6c nhHc i nhHc l/i ngy cng

khHt khe hQn, nhanh hQn v p lucthnangc ln HS cuthngang m/nh hQn; p lucthnangc trong

luy,n tVp s| c4ng thng hQn trong bi tVp thucthnangc hnh. Tuy nhin, p lucthnangc

khng nn qu cao m chc vucthhuyena uthhoi khuyn khch HS lm bi ch:u kh

hQn. Th%i gian cho luy,n tVp, thucthnangc hnh cuthngang khng nn ko di qu d `

gy nn sucthnang nh/t nh|o v nhm chn. Cn thit k cc bi tVp c sucthnang phn

ho khuyn khch mDi i tucth6ng HS >u tham gia thucthnangc hnh luy,n

tVp ph h6p v"i n4ng lucthnangc cuthhoia mnh. Cuthngang c th t@ chucthsacc cc ho/t ng

luy,n tVp, thucthnangc hnh thng qua nhi>u ho/t ng khc nhau, k c vi,c t@

chucthsacc thnh cc tr chQi hDc tVp nhbm lm cho HS ho hucthsacng hQn, 0ng

th%i qua cc ho/t ng , cc ka n4ng cuthhoia HS cuthngang ucth6c rn luy,n.


6. V d minh ho


Khi hDc bi Cng thucthsacc nghi,m cuthhoia phucthQng trnh bVc hai (?/i s 9),

HS cn ucth6c luy,n tVp :

+ Xc :nh ng cc h, s a, b, c cuthhoia phucthQng trnh;

+ Thnh th/o vi,c tnh bi,t thucthsacc ;

+ Nh" v vVn duthnangng thnh th/o cng thucthsacc nghi,m xc :nh nghi,m cuthhoia

phucthQng trnh bVc hai.

Sau HS thucthnangc hnh gii cc phucthQng trnh bVc hai v"i cc [n khc

nhau, gii cc phucthQng trnh m sau qu trnh bin @i m"i uctha ucth6c v>

phucthQng trnh bVc hai...

V duthnang minh ho qua mn m nhc:

D/y mt bi ht:

+ GV d/y tucthhuyenng cu ngHn (lm mYu qua ting n hay giDng ht).

+ HS ht theo (thucthnangc hnh).

+ Sau khi d/y xong c bi ht, GV cho HS tVp g ,m, ht kt h6p vVn

ng, ht kt h6p tr chQi hay tVp biu di`n... ? chnh l nhucthngang khu

luy,n tVp cui cng cuthhoing c bi hDc gip HS hnh thnh ka n4ng ht

(bao g0m cch ht, hDc thuc bi ht v ht ng...).


Khi d/y hDc bi So snh (Ngucthnga v4n 6 tVp 2, Bi 19 v 20), GV chDn mt cu

no c hi,n tucth6ng so snh ngang bbng v so snh hQn km lm mYu.

Sau khi phn tch, HS hiu v nHm vucthngang mYu, HS tucthnang mnh t/o ra cc

cu khc nhau theo mYu so snh theo yu cu cuthhoia GV cho n khi hnh

thnh ucth6c ka n4ng.

V duthnang minh ho qua mn Ti:ng Anh:

HS E u c=p THCS c th ucth6c nghe cc mYu i tho/i chucthsaca c=u trc

cu thng thucth%ng nhucth h8i v tr l%i v> th%i tit, v duthnang: Whats the weather

like? Its cold (hot/sunny/rainy). HS cn phi ucth6c GV lm r nghaa

(bbng gii thch, cho v duthnang hoc thVm ch phi d:ch sang ting Vi,t nu


c=u trc cu khng c trong ting m ., v duthnang: /i tucthhuyen it dng chc

th%i tit) v hiu ucth6c cch sucthhoi duthnangng c=u trc cu, cch pht m, ngucthnga i,u

cu h8i (xung giDng). HS c th vVn duthnangng h8itr l%i v> th%i tit trong

cc tnh hung g6i (v duthnang: cc tranh v| tr%i nng/l/nh/=m) hoc

trong tnh hung thVt E cc :a danh khc nhau ducthnanga vo bn tin ducthnang bo

th%i tit trn ti vi; v duthnang: Whats the weather like in Hanoi/Hue/Ho Chi

Minh City? Its......


dy hc luyn tp v thc hnh



dy hc luyn tp v thc hnh

GV > xu=t mt v duthnang (mt bi d/y) v> PPDH luy,n tVp v thucthnangc hnh

trong mn hDc m mnh ang ging d/y.


Hot ng 4. Tho lun nhm v phng php dy hc luyn

tp v thc hnh v cc v d xut Hot ng 3

G*i :






no khc,


GV tucthnang rt ra nhucthngang ucthu nhucth6c im chnh v cch sucthhoi duthnangng PPDH luy,n

tVp v thucthnangc hnh trong mn hDc cuthhoia mnh nhbm /t hi,u qu cao nh=t.


Ho/t ng 2 trn.

D


Ni dung 7

TM HIU V PHNG PHP DY HC BNG BN T DUY

Hot ng 1. Tm hiu v phng php dy hc bng bn t duy

NHIM V

B/n hy Dc nhucthngang thng tin cQ bn cuthhoia Ho/t ng 1 lm r:

1. Bn ch=t cuthhoia phucthQng php d/y hDc bbng bn 0 tucth duy v quy trnh thucthnangc

hi,n n.

2. Chc ra nhucthngang ucthu im, nhucthngang h/n ch v nhucthngang im cn lucthu v> phucthQng

php d/y hDc bbng bn 0 tucth duy.


THNG TIN C BN

1. Bn cht

B;n < tucth duy (Mindmap), cn gDi l sQ 0 tucth duy, lucth6c 0 tucth duy: l

PPDH ch trDng n cQ ch ghi nh", d/y cch hDc, cch tucthnang hDc nhbm tm

ti, o su, mE rng mt tucthEng, h, thng ho mt chuthhoi > hay mt

m/ch kin thucthsacc, bbng cch kt h6p vi,c sucthhoi duthnangng 0ng th%i hnh nh,

ucth%ng nt, mu sHc, chucthnga vit v"i sucthnang tucth duy tch cucthnangc. Bn 1 tucth duy gip

th/ hin ra bn ngoi cch thucthsacc m no b& chng ta hot &ng.

HS tucthnang ghi chp kin thucthsacc trn bn 0 tucth duy bbng tucthhuyen kho v chnh,

cuthnangm tucthhuyen vit tHt v cc ucth%ng lin kt, ghi ch, bbng cc mu sHc, hnh

nh v chucthnga vit. Khi tucthnang ghi theo cch hiu cuthhoia chnh mnh, HS s| chuthhoi

ng hQn, tch cucthnangc hDc tVp v ghi nh" b>n vucthngang hQn, d` mE rng, o

su tucthEng. Mii ngucth@i ghi theo m&t cch khc nhau, ghi theo cch hi/u

cuthhoia mnh, khng rp khun, my mc. Ti/m mnh cuthhoia bn 1 tucth duy l

kch thch hucthsacng th v to cm hucthsacng sng to.

PPDH bbng bn 0 tucth duy l PPDH m GV, HS thucthnangc hi,n nhi,m vuthnang

d/y hDc thng qua vi,c lVp bn 0 tucth duy. Sucthhoi duthnangng PPDH bbng bn 0

tucth duy trong d/y kin thucthsacc m"i, n tVp, cuthhoing c, h, thng ho v kim

tra tri thucthsacc.

PPDH bbng bn 0 tucth duy l PPDH m GV t@ chucthsacc cc ho/t ng cho HS

lVp bn 0 tucth duy thucthnangc hi,n nhi,m vuthnang hDc tVp trong qu trnh hDc tVp.


PPDH bbng bn 0 tucth duy l phucthQng php t@ chucthsacc cho HS tm hiu mt

v=n >, thucthnangc hi,n mt nhi,m vuthnang hDc tVp thng qua vi,c lVp bn 0 tucth duy

(cc bn 0 tucth duy chuthhoi yu do HS thit lVp trong qu trnh hDc tVp, h/n

ch vi,c sucthhoi duthnangng cc bn 0 tucth duy c syn).

Sucthhoi duthnangng PPDH bbng bn 0 tucth duy trong d/y kin thucthsacc m"i, n tVp,

cuthhoing c, h, thng ho v kim tra tri thucthsacc.

Trong PPDH ny HS tucthnang mnh thit lVp bn 0 tucth duy v> kin thucthsacc nghaa l

tucthnang mnh v|, vit, th hi,n ra bn ngoi nhucthngang suy ngha, hiu bit cuthhoia

mnh v> kin thucthsacc bi hDc bbng bn 0 tucth duy, thng qua chim

lanh kin thucthsacc. GV l ngucth%i c v=n, trDng ti, t@ chucthsacc cho HS cc ho/t

ng hDc tVp.

2. Quy trnh thc hin

Bucthc 1 : LVp bn 0 tucth duy

HS lVp bn 0 tucth duy theo nhm hoc c nhn v"i cc g6i lin quan

n chuthhoi > kin thucthsacc cuthhoia bi hDc.

I) Chn tucthhuyen trung tm (hay cn gDi l tucthhuyen kho, keyword) l tn cuthhoia mt bi hay

mt chuthhoi > hay mt ni dung kin thucthsacc cn khai thc (cuthnangm tucthhuyen tnh trung

thucthnangc, tucthhuyen Qn, tucthhuyen ghp, d=u hi,u chia ht, hnh chucthnga nhVt,...) hoc l

mt hnh nh, hnh v| m ta cn pht trin (hnh vung, hnh thoi,).

V duthnang: Thit lVp bn 0 tucth duy bi Hnh chucthnga nhVt Ton 8.

BHt u bbng cuthnangm tucthhuyen trung tm Hnh chucthnga nhVt

hoc l mt hnh v| hnh chucthnga nhVt

B

D

C

A


II) Vj nhnh cIp 1

Cc nhnh c=p 1 chnh l cc ni dung chnh cuthhoia bi hDc hay chuthhoi >

(hay tn cc muthnangc cuthhoia sch gio khoa), chng h/n nhucth v"i bi Hnh chucthnga

nhVt c 3 muthnangc l: :nh nghaa, tnh ch=t, d=u hi,u nhVn bit, tuy nhin

nn thit lVp bn 0 tucth duy c 4 nhnh c=p 1, thm nhnh cc hnh

trong thucthnangc t c d/ng hnh chucthnga nhVt.

Cc nhnh c=p 1 khng phi hon ton ducthnanga vo cc > muthnangc cuthhoia SGK,

chng h/n bi Gin d: (Gio duthnangc Cng dn 6), mc du SGK khng

c cc muthnangc r rng nhucthng ta c th chDn lDc ni dung chnh c th

v| 4 nhnh c=p 1, l: 1) K tn gucthQng nhucthngang ngucth%i sng gin d: m

em bit; 2) Biu hi,n sng gin d:; 3) Biu hi,n tri gin d:; 4) K ho/ch

rn luy,n.


III) Vj nhnh cIp 2, 3,...

Cc nhnh con c=p 2, 3, chnh l cc nhnh con cuthhoia nhnh con tructh"c

(hay ni r hQn nhnh con c=p 2, 3, l cc trin khai cuthhoia nhnh tructh"c ).

Chng h/n, nhnh c=p 1 d=u hi,u nhVn bit (bi Hnh chucthnga nhVt) c 4

nhnh con c=p 2, mWi nhnh l mt d=u hi,u.

Bucthc 2: Bo co, thuyt minh bn 0 tucth duy (vucthhuyena thit lVp E bucth"c 1)

Cc cuthnangm tucthhuyen, cng thucthsacc, hnh v|,

Module THCS 18

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