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THPT CHI LNGN TN THPT 2011

Tun 1, Tit 1 - 8KHO ST HM S V PTTT TI MT IM

I. Mc ch - yu cu :- Gip hc sinh thc hin kho st v v th ca cc hm s:

+ y = ax3 + bx2 + cx + d (a 0),

+ y = ax4 + bx2 + c (a 0)

+ax b

ycx d

+=

+(ac 0), trong a, b, c, d l s thc.

- Vit c PTTT ti mt im cho trc : honh 0x , tung 0y hoc h s gc k.

Vit phng trnh tip tuyn ca (C): y = f(x) tiM0(x0;y0) (C).

Bc 1: Tm cc thnh phn cha c x0, y0, f(x0) bi cc h thclin h :

0 0( ) y f x = hoc 0'( )f x k= (c gi l h s gc ca tip

tuyn)

Bc 2: Nu dng pttt : y y0 = f(x0) ( )0x x hay y y0 = k(x x0)(*).

Thay cc s liu x0, y0, f(x0) vo (*) v rt gn ta c kt qu- Bit cch da vo th bin lun theo m s nghim ca phng trnh.

II, Ni dung n tp :

1. Bin php n tp

- Cho hc sinh vn dng cc bi tp quen thuc n gin, t khi qut

kin thc l thuyt v khc su c cc khi nim lin quan n hm s.

2 . Ni dung bi tp :

Bi 1: Cho hm s 3 21 3 54 2

y x x = - +

a) Kho st v v th (C) ca hm s cho.

b) Vit PTTT ca th (C) ti im c honh bng -2.

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c) Vit PTTT ca th (C) bit rng tip tuyn song song vi ng thng

93

4y x= - +

d) nh m phng trnh 3 21 3 5 2 04 2

x x m- + - = c ng 1 nghim

Bi 2: Cho hm s 3 23 y x x = - +

a) Kho st v v th (C) ca hm s.

b) Vit PTTT ca th (C) ti im c tung bng 4.

c) Tnh din tch hnh phng c gii hn bi (C ) , trc 0x v x=1

Bi 3: Cho hm s 4 22 2 y x x = - +a) Kho st v v th (C) ca hm s.

b) Vit PTTT ca th (C) ti im c honh l nghim ca phng trnh

''( ) 02

xy =

Bi 4: Cho hm s 4 21 2 44

y x x = - + - .


b) Vit PTTT ca th (C) ti im c honh bng 3.

Bi 5 : Cho hm s y = 32 1

x

x

+

.


b) Lp pttt ca th (C) bit rng tip tuyn song song vi ng thng y=x

Bi 6 : Cho hm s y = 2 41xx ++ .


b) Lp pttt ca th (C) ti im c tung bng 1.

Bi 7: Cho hm s 4 23 1 y x x = + + c th l (C).

a) Kho st v v th (C).Trang 2

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b) Vit pttt ca th (C) ti im c tung bng 5.

c) Tnh din tch hnh phng c gii hn bi (C) v ng y=5

Bi 8: Cho hm s y = 2 4

1

x

x

+

+.


b) Gi D l phn hnh phng c gii hn bi (C ), 0x v 0y. Tnh th

tch khi trn xoay khi D quay quanh trc 0x

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3. Bi tp t ng t

Bi 1: Cho hm s y = x3 3x + 5.



c) Vit PTTT ca th (C) song song vi ng thng 9 2011y x= - .

Bi 2: Cho hm s y = (x -1)2(4-x)


b) Vit PTTT ca th (C), bit rng h s gc ca tip tuyn bng -9.

Bi 3: Cho hm s y = (x -1)2(x+1)2.


b) Vit PTTT ca th (C) ti im c tung bng 2.

Bi 4: Cho hm s 4 21 332 2

y x x = - + .


b) Vit PTTT ca th (C) ti im c honh l nghim ca pt ''(2 ) 0f x = .

Bi 5 : Cho hm s y = 22

xx

+ .


b) Lp pttt ca th (C) ca hm s ti cc giao im vi trc tung v trc

honh.

Bi 6 : Cho hm s y = 2 11

x

x

+

.


b) Lp pttt ca th (C) ti im c honh bng -1.

4. nh gi Rt kinh nghim tit dy :

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Tun 2, Tit 1 - 8

BI TP TNG HP* HM S TRNG PHNG:

Cu 1 : Cho hm s : =y 4 21

24

x x

1) Kho st s bin thin v v th (C) ca hm s cho.2) Vit phng trnh tip tuyn ca (C) ti im c honh bng 23) Vit phng trnh tip tuyn ca (C) bit tip tuyn vung gc vi trc tung.

Cu 2: Cho hm s y = f(x) = - x4 2(m 1)x2 + 2m 11) nh m th hm s ct trc honh ti 3 im phn bit.

2) Kho st s bin thin v v th (C) ca hm s khi m = 0.3) Vit phng trnh tip tuyn ca (C) bit honh ca tip im bng 34) Vit phng trnh tip tuyn ca (C) bit tip tuyn vung gc vi ng thng y =1 1

24 12x +

Cu 3: Cho hm s y = ( 2 x2 )2 C th (C) .1) kho st v th ( C ) ca hm s .2) Da vo th ( C ) , bin lun theo m s nghim ca : x4 -4x2 m = 03) Gi A l giao im ca ( C ) v Ox , xA > 0 . Vit phtrnh tip tuyn vi ( C ) ti im A.

Cu 4: Cho hm s 4 22 1 = x xy c th (C)

a) Kho st s bin thin v v th (C).b) Vit phng trnh tip tuyn ca (C) bit tung ca tip tuyn bng -1.

Cu 5: Cho hm s4

2 3

2 2

xy x= +


b) Vit phng trnh tip tuyn ca (C) ti giao im ca (C) vi y =4

2 94x2 2

xx +

Cu 6: Cho hm s y = (1 m) x4 mx2 m 1 (Cm)a) Kho st v v th (C) khi m = 2..

b) Tm m hm s (Cm) c ng mt cc tr.c) Tm m hm s t cc tiu ti x = 1.

Cu 7: Cho hm s 4 2y x 2x= - + +1. Kho st s bin thin v v th (C) ca hm s cho.2. Bin lun theo m s nghim ca phng trnh 4 2x 2x 1 m 0- - + =3. Vit PTTT ca th hm s ti trung im ca on thng ni 2 im cc tr ca hm s.Cu 8: Cho hm s y = x4 + mx2 m 1 (Cm)

1) Kho st v v th hm s khi m = -2

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2) Vit phng trnh tip tuyn ca (C) bit tip tuyn song song vi trc honh.3*) Tm im c nh A c honh dng ca (Cm). Vit phng trnh tip tuyn ti A son

song vi ng thng y = 2x.

* HM S PHN THCax

x+d

by

c

+=

Cu 1: Cho hm s4

4=

y

x(C)

1) Kho st s bin thin v v th (C).2) Vit phng trnh tip tuyn (d) ca (C) ti im thuc (C) c honh l 3.3) Tm din tch hnh phng gii hn bi (C), tip tuyn (d) v trc Oy.4) Bin lun theo k s giao im ca (C) v ng thng ( ) i qua A(-4, 0), c h s gc k.

Cu 2: Cho hm s : y=3 2

1

+

x

x

1) Kho st s bin thin v v th (C) ca hm s cho .

2) Chng minh rng ng thng y = -2x-m lun ct (C) ti hai im phn bit.3) Vit phng trnh tip tuyn ca (C) ti giao im ca (C) vi trc tung.

Cu 3: Cho hm s :3 2

.1

=

x

yx

1) Kho st s bin thin v v th hm s cho.2) Tm tt c cc gi tr ca tham s m ng thng y= mx+2 ct th hm s cho t

hai im phn bit.3) Vit PTTT ca (C) ti giao im ca (C) vi trc honh.

Cu 4: Cho hm s:2 1

1

=

xy

x

c th (C)

1) Kho st s bin thin v v th (C).2) Vit pt tip tuyn vi (C) bit tip tuyn vung gc vi t (d): 12x + 3y + 2 = 0.3) Tm to giao im ca (C) vi ng thng 2x 3y 4 = 0.

Cu 5: Cho hm s y =1

1

+

x

x

1. Kho st s bin thin v v th (C) ca hm s.2.Vit phng trnh tip tuyn vi (C) ti im thuc (C) c honh x0 = -23.Gi (H) l hnh phng gii hn bi (C) v 2 trc ta . Tnh din tch hnh phng (H).

Cu 6: Cho hm s2x 1

yx 2

+=

-1. Kho st s bin thin v v th (C) ca hm s cho.2. Vit phng trnh tip tuyn ca th (C) ti im trn (C) c tung y 3=- .

Cu 7: Cho hm s2 1

1

+=

x

yx

1 Kho st s bin thin v v th (C) ca hm s.2. Tm tt c cc gi tr ca tham s m ng thng y = (m2 + 2)x + m song song vi tip tuyca th (C) ti giao im ca th (C) vi trc tung.

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* HM S BC BA:Cu 1: Cho hm s 3 3= + y x x c th (C)

1. Kho st v v th (C)2. Dng (C) bin lun theo m s nghim phng trnh 3 3 0 + = x x m

3. Vit phng trnh tip tuyn ca (C) vung gc vi ng thng (d): x 9y + 3 = 0Cu 2: Cho hm s 3 23 1= + + y x x c th (C)

1. Kho st v v th (C).2. Vit phng trnh tip tuyn ca th (C) ti A(3;1).3. Dng th (C) nh k phng trnh sau c ng 3 nghim phn bit 3 23 0 + = x x k

Cu 3: Cho hm s y = 3x2 x3 c th l ( C).1. Kho st s bin thin v v th ( C) ca hm s.2. Vit phng trnh tip tuyn vi ( C) ti im A thuc ( C) c honh x0 = 3.

Cu 4: Cho hm s 3 26 9= + y x x x , c th (C)1. Kho st s bin thin v v th (C) ca hm s.

2. Tnh din tch hnh phng gii hn bi th (C) v ng thng y = x.

* THAM S M:

Cu 1: Cho hm s y =2x

x

+. Chng minh rng (C) lun ct d: y = 2x + m ti hai im phn bit (

0)Cu 2: Cho hm s y = x4 4x2 + 4 c th (C)

a, Kho st v v th (C).B, Tm m ng thng y = m ct (C) ti 4 im phn bit.

Cu 3: Cho hm s y = mx4 + (m2 9).x2 + 10. Tm m hm s c 3 cc tr.

Cu 4: Cho hm s y = x3 3mx2 + 3(2m -1)x + 1. Xc nh m hm s c 1 cc i v 1 cc tiuTm to im cc tiu.Cu 5: 3 23 3 1 y mx mx x= + . Xc nh m hm s c hai cc tr nm v hai pha trc tung.Cu 6: Tm m th hm s y = (x 1) (x2 + mx + m) ct trc hanh ti ba im phn bit.Cu 7 : nh m y= ( ( 1133 2223 + mxmmxx t cc i ti x=1. S :m=2

Cu 8 : Cho hm s y= baxx

+ 24

2. nh a,b hm s t cc tr bng 2 ti x=1

Cu 9 : Cho hm s y=1

2

++

x

mxxnh m hm s c cc tr v 2 gi tr cc tr cng du.

Cu 10: nh m hm s: y= x3

3mx2

+ (m+2)x m ng bin trn

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Tun 3, tit 1 8 HM S M LGARITI, Mc ch yu cu:

- Hc sinh nm c cc cng thc lin quan n hm s m, lgarit. Cng thc o hm ca hms m v hm s lgarit.- Bit c phng trnh m, phng trnh lgarit n gin nht.- p dng cc cng thc bin i v a phng trnh cho v phng trnh n gin.- Gii c mt s bt phng trnh m v lgarit n gin.II, Ni dung n tp:1, Bin php thc hin:- Tm tt kin thc l thuyt thng qua cu hi lin quan n cng thc.- Hc sinh vn dng da theo s hng dn ca hc sinh.- Gio vin hng dn hc sinh cch hc cng thc, cch nh, phng php tng qut phn tch bi

ton gii phng trnh.2, Bi tp:DNG 1. PHNG TRNH M

1) 2 8 1 32 4 + = x x x

2)22 3 3 53 9 + = x x x

3) 22 2 20x x ++ =4) 1 2 3 1 23 3 3 9.5 5 5+ + + + ++ + = + + x x x x x x

5)2 3 3 7

7 11

11 7

=

x x

6)2 5 4

14

2

+ =

x x

7) 1 13 3 10+ + =x x 8) x x3.9 2.9 5 0 + =9) x x2.16 17.4 8 0 + =10) 6x 3xe 3e 2 0 + =11) 4 8 2 53 4.3 27 0+ + + =x x

12)3.25x + 2. 49x = 5. 35x

13) ( ) ( )2 3 2 3 14x x

+ + =

DNG 2. PHNG TRNH LGARIT1) ( )2log .( 1) 1x x =

2) ( )3 3log log 2 1+ + =x x 3) log(x - 2) + log(x -3) = 1 - log54)

( ) ( )4 4 4log x 2 log x 2 2 log 6+ =5) lnx + ln(x+1) = 0

6)1 2

14 ln 2 ln

+ = +x x

7) log2

3 1

2

3.log 2 0+ + =x x

8) 2 32 2log log 4 0x x+ =9) 94log log 3 3+ =xx 10)log4(x +2 ) = log2x

11) ( )5 15

log 1 log ( 2) 0x x + =

12) x x 15 25log (5 1).log (5 5) + =

DNG 3: BT PHNG TRNH M - LGARIT

1) x 416 82)

2 51

93

+ x x

4)24 15 4

3 412 22

+ 3. 5x

6) 2 3 7 3 16 2 .3 x x x+ + +

8) log2( x2 4x 5) < 49) 2log8( x- 2) log8( x- 3) > 2/3

10) 13

3 1log 1

2

>

+x

x

11) 22 2

log log x 0x +

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12) 2 2xlog x log 8 4+

13)1 1

11 log log

+ > x x

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Gio n n thi TN THPT 2011GV: Phan Minh Tm

DNG 4: Rt gn biu thcBi 1: Vit di dng lu tha s m hu t: a) 5 3 222 b) 4 32 xx (x > 0 )Bi 2: Rt gn

A = axa 1(4

1x 1 )(

11

11

11

11

+

+

xa

xa

xa

xa) (ax 0; x a ) B = 11log

5

1 +

121log 25 - 1211

log 5

Bi 3: Tm 32log 49 bi t 14log 2 = a

Dng 5:MT S DNG TON KHC

Bi 1: Cho hm s 2x xy e += . Gii phng trnh y y 2y 0 + + =

Bi 2: Tnh o hm: a) y = )532(log 25 xx b) y = 3ex- 5 sin3x + ln(x+1)

3, nh gi Rt kinh nghim:

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Tun 4, Tit 1 8 NGUYN HM TCH PHNI, MC CH YU CU:- Hc sinh nm vng cng thc tnh nguyn hm v p dng tm nguyn hm ca cc hm s.- Tm nguyn hm ca hm s tha mn iu kin cho trc.- Hc sinh nm c cng thc Niutn Lepnit.- Tnh c tch phn ca cc hm s s cp.- Nhn bit v vn dng chnh xc phng php tch phn i bin s.- Nhn bit v vn dng chnh xc phng php tch phn tng phn.- Nm vng cng thc tnh din tch hnh phng, th tch vt th trn xoay.- Vn dng tnh din tch v th tch c bn.

II, NI DUNG N TP:1. Phng php thc hin:- Chun b h thng bi tp c bn lin quan n cng thc o hm v nguyn hm, trn c svn dng gip hc sinh h thng li cc cng thc v thuc c cng thc o hm v ng hm

- Nu phng php nhn dng chung cho hc sinh cc em d dng tip cn v gii c biton tm nguyn hm, tch phn.- Nu mt s sai lm thng gp trong gii ton tch phn v hng khc phc.2. Bi tp:* NGUYN HM

Bi 1. Tm nguyn hm F(x) ca hm s ( ) 3 2 5f x x x = - + bit ( )2 5F =

S: ( )4

2 5 54

xF x x x = - + - .

Bi 2. Tm nguyn hm F(x) ca hm s ( ) 1f x xx

= + bit ( )2

2eF e =

S: ( )2

ln 12

xF x x= + - .

Bi 3. Tm nguyn hm F(x) ca hm s ( ) 21

sincos

f x xx

= + bit2

4 2F

p =

S: ( ) cos tan 2F x x x = - + + - .

Bi 4. Tm nguyen ham F(x) cua f(x)= x3

x2

+2x1 biet rang F(0) = 4.Kt qu: F(x) =

34

34 xx +x2x+4

Bi 5. Tm A va B sao cho vi moi x 1 va x 2 , ta co:

1223

12

+

=+

+x

B

x

A

xx

x

T o, hay tm ho nguyen ham cua ham so:23

1)(

2 ++

=xx

xxf

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Kq: A=3; B= 2F(x) = 3lnx22lnx1+ C= ln 23

)1(

2

x

x

+ C

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* TCH PHN

Bi 1. Tnh cc tch phn sau:

( )1

4 2

0

3 2I x x dx = - +2 3 2

2

1

2 5x xI dx

x

-=

1

0

2 3

2

xI dx

x

+=

-

( )1

53 4

0

1I x x dx = -2

1

0

xI xe dx -= 2

2

0

sin cosI x xdx

p

=

2

0

sin

1 cos

xI dx

x

p

=+

2

0

3 lnxI dx

x

p

+=

6

2

1

3I x x dx = +

2 2

1

1 3lnln

xI xdx

x

+=

6

0

1 4sin cosI x xdx

p

= +

82

0

cos 2I xdx

p

=

I = dxxe xx )1.(

2

1

22

I = 1

0

200932 )1( dxxx I = +6 3lne

e

dxx

x

I =( )

6

3

0 2 1

cos

sin

xdx

x

+(3/16) I =

( )1 3 2ln

edx

x x+ I =

+

+1

02 44

32dx

xx

x

I = 2

4ln

e

exx

dx (s: 7/24) I =

4

0

4

5

1ln

x

xdx

Bi 2. Tnh cc tch phn sau:

2

0

cosI x xdx

p

= 1

0

xI xe dx = 1

lne

I x xdx = I = +2

1

2

)1ln(

x

dxx

( )1

2

0

1 xI x e dx = +4

0

2 cos2I x xdx

p

= ( )

3

2

2 ln 1I x x dx = - I = +2

1

2 .ln)1( dxxx I=

+1

0

2 .).1( dxex x ;

Bi 3: Tnh cc tch phn:

( )

2

0

2sin 3 cosI x xdx

p

= + ( )2

1

0

2 xI x xe dx = + ( )2

sin

0

1 cosxI e xdx

p

= +

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I = +2

0

2 cosxdx)sin(

xx I= +2

1

.).ln( dxxxex I= +2

1

.).ln( dxxxex

I= e

xdxx1

2ln)1( I = +

2

1

.).ln( dxxxex

* NG DNG:Bi 1. Tnh din tch hnh phng gii hn bi:a, y = x2 + 4x v trc honh. (s: 32/3)

b, y = x2 2x + 3 v y = x + 1 ; x = 0; x = 2.(s:1)

c, y = ex; y = 2 v ng thng x = 1.

d, y = x3 x ; y = x x2 . (s: 37/12)

Bi 2. Tnh th tch vt th trn xoay:a, khi cho hnh phng to bi y = 1 x2; y = 0 khi quay quanh trc Ox.

S: 15

16

b, khi cho hnh phng to bi y = tanx; y = 0, x = 0, x =4

quay quanh trc Ox.

S:

41

Bi 3: Tnh din tch hnh phng y = 2x2 3x v y = 3x2 + 4 v x = -4; x = 2.Bi 4: Tnh din tch hnh phng y = x3 v y = - 2x2 + 3; x = -2; x = 0.

Bi 5: Tnh din tch hnh phng y = 3x2 3x v y = 2x2 - 2 v x = -1; x = 3.Bi 6: Tnh din tch hnh phng gii hn bi y = 2x3 v y = x2 + 1; x = 0; x = 2.Bi 7: Tnh th tch vt th trn xoay khi cho hnh phng to bi y = cosx; y = 0; x = 0; x=quay quanh trc Ox.

Bi 8: Tnh th tch vt th trn xoay khi cho hnh phng to bi y = 3 21

3 y x x = - ; y = 0;

x = 0; x = 3 quay quanh trc Ox.

* Rt kinh nghim tit dy:

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Tun 5, Tit 1 8 PHNG PHP TO TRONG KHNG GIANTit 1, 2 CC PHP TON VCT

I. Mc ch - yu cu:Tm to ca vct, di vct, to trung im, trng tm tam gic, chng minh

vct cng phng, tnh cht hai vct bng nhau; tch v hng, c hng ca hai vct.II. Chun b:

Gio n, h thng kin thc l thuyt, bi tp vn dng, bi tp rn luyn.III. Ni dung bi dy:

H thng kin thc l thuyt (1):Gi cho hc sinh nhc li cch tnh to vct; di vct; to trung im ca

on thng; trng tm ca tam gic ABC.PHNG PHP NI DUNG

Gi cc em ln thchin tnh theo yu cu

bton...

Bi tp 1: Cho 4 im trong khng gian Oxyz ln lt c to l:A(1;2;-1); B(2;1;3);C(1;-2;1);D(0;1;1)

a, Tm ta ca cc vct ;;;; ADBCACABb, Tm di ca cc vct ;c, Xc nh ta trung im ca AB, AC, BC, BD;d, Xc nh ta trng tm ca tam gic ABC, ACD; ADB; BCD;e, Tm ta im M t gic ABCM l hnh bnh hnh.

? iu kin t gicABCM l hnh bnhhnh?

Gii:a, ( )1; 1;4AB = ; AC = (0; -4; 2); BC= (-1; -3; -2); DA = (-1; -1; 2)

b, di cc vct trn ln lt l: 18 ; 20 ; 14 ; 6c, Trung im ca cc on trn l: I(....

d, Trng tm tam gic l:e, Gi M(x;y;z) l im cn tm, khi ABCM l hbh khi v chkhi

CM BA= ; iu ny xy ra khi

1 1

2 1

1 4

x

y

z

= + = =

hay x = 0; y = -1; z = -3

Vy M(0; -1; -3)Gi cch tnh, hcsinh vn dng.

Bi tp 2: Trong khng gian, cho cc vct:3 ; 2 ; 2a i k b i j k c i j k = + = = +

a, Hy tnh ; .( )b c a b c+ + ; ( ); .a b a b c b, Tnh c ; ba +

Gii: hc sinh gii; gio vin sa cch trnh by...S dng tnh cht haivct bng nhau tmcc nh cn li cahnh hp.

Bi tp 3: Trong khng gian vi h to Oxyz, cho hnh hpch nht ABCD.ABCD bit to im A(1;0;1); B(2;1;2);D(1; -1; 1); C(4;5;-5). Tm ta cc nh cn li ca hnh hp.Gii:

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* Cng c kin thc:Gio vin nhc li cc kin thc trng tm ca tit dy: Tm to vct; tnh cht hai

vct bng nhau v ng dng; tm to trung im, trng tm tam gic; tm di onthng; tch v hng , c hng ca hai vct.* Dn d: V gii cc bi tp n v gii tip cc bi tp tng t sau:

BI TP RN LUYN TIT 1-2----***----

BI 1: Trong khng gian Oxyz, cho im A(0; 2; -1) ; B(1;1;3); C(-1; 2; -2)a, Chng minh ABC lp thnh tam gic. Tm to trung im ca cc cnh tam gic

ABC, trng tm G ca tam gic ABC;b, Tm di 3 cnh ca tam gic ABC.c, Tnh din tch tam gic ABC.

BI 2: Cho 3 im A(1;0;0); B(0; 0; 1) ; C(2;1;1)a, Tm khong cch gia 2 im A v B; A v C; B v C.

b, Tnh s o cc gc trong tam gic ABC.c, Tm chu vi v din tch tam gic ABC.

BI 3: Tm im M trn mt phng Oxz sao cho n cch u 3 im A(1;1;1); B(-1;1;0);C(3;1;-1)

BI 4: Cho im A(1; 2; 1); B(5;3;4); C(8;3; 0)a, CMR tam gic ABC vung.

b, Tnh din tch tam gic ABC.c, Tm im D ABCD l hnh ch nht.

BI 5: Cho A(1; 0; 0) ; B(0; 1; 0) ; C(0; 0; 1); D(-2; 1; -1)

a, Chng minh rng, 4 im A,B,C,D lp thnh t din.b, Tnh gc to bi cc cp cnh i din ca t din.c, Tnh th tch t din ABCD v di ng cao h t nh A.

BI 6: Cho A(1; -2; 2) ; B(1; 4; 0); C(-4; 1; 1) ; D(-5; -5; 3)a, Chng minh rng AC vung gc BD.

b, Tnh din tch ABCD.-----------------------------

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Tit 3,4 - TUN 5PHNG TRNH MT PHNG

I. Mc ch - yu cu:+ Hc sinh nm c dng phng trnh mt phng trong khng gian, xc nh c

yu t trong phng trnh mt phng.+ Hc sinh vit c phng trnh mt phng n gin.+ Lm quen mt s dng ton s dng tch c hng tm VTPT ca mt phng.

II. Chun b:H thng l thuyt, bi tp hc sinh thc hin ti lp, bi tp rn luyn ti nh.

III. Ni dung bi dy:H thng kin thc l thuyt:

* Kin thc c bn:+ Phng trnh mp (P): Ax + By + Cz + D = 0 c VTPT l ( ); ;n A B C = v i qua im

M...+ Nu mp(P) c VTPT l ( ); ;n A B C = v i qua im M(xo;yo;zo) th (P) c dng:

A(x - xo) + B(y - yo) + C(z - zo) = 0+ Nu (P) c cp vc t ;a b c gi song song hoc trng (P) th VTPT l n a b=

* Ch :+ Hai mt phng song song th vtpt ca mp ny cng l vtpt ca mp kia.+ ng thng vung gc mp th vtpt ca mp cng l vtcp ca ng thng.+ Mt phng (P) tip xc vi mt cu (S) tm I ti tip im H th mt phng (P) c

vct php tuyn l vctIH .

PHNG PHP NI DUNGHc sinh nm vng tnh

cht song song v vunggc gia mp vi mp; giamp vi t

Bi 1: Cho im M(2; -1; 3) v mt phng (P): 2x - y + 3z - 1 =

0; ng thng d: 1 22 3 1

x y z += =

a, Vit phng trnh mt phng i qua im M v song song vi(P).

b, Vit phng trnh mt phng i qua im M v vung gc t d.c, Vit phng trnh mt phng i qua O, vung gc mp(P) vsong song dGii:

S dng tch c hng cahai vct tm vtpt

Bi 2: Cho im A(2; 3; -4); B(4; -1; 0) ; C(2; 0; -3)

a, Vit phng trnh mt phng i qua im A v vung gc viBC.

b, Vit phng trnh mt phng i qua A, B v song song vi OC.c, Vit phng trnh mt phng i qua 3 im A, B, C.d, Vit phng trnh mt phng trung trc ca AB.e, Lp phng trnh mt phng cha trc Oy v song song AB.Gii:Bi 3: Cho mt cu (S): x2 + y2 + z2 - 2x + 4y - 6z - 2 = 0

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Kt hp vi iu kin tipxc ca mt cu vi mt

phng.

a, Vit phng trnh mt phng tip xc vi mt cu (S) ti M(1;-2; -1)

b, Vit phng trnh mt phng song song vi mt phng(P): 2x - y + 2z -1 = 0 v tip xc vi (S).

* Cng c: + Dng pt mp+ xc nh yu t nh vtpt ca mp; vtcp ca t; tm v bn knh ca mt cu.+ Cc dng ton vit pt ca mt phng thng gp.

* Dn d: V gii li cc bi tp trn v hc k l thuyt c bn v gii thm mt s bi tp sau:BI TP RN LUYN VIT PHNG TRNH MT PHNG

Bi 1: TRONG MT PHNG OXYZ, Vit phng trnh mt phng:

a, i qua im M(1; 2; 3) v vung gc vi ng thng d:1 3

1 2 3

x y z = =

.

b, i qua im N(1; 3; 5) v vung gc trc Oz.

c, i qua gc to v vung gc vi ng thng

2 3

: 3

1 2

x t

y t

z t

=

= = +d, Qua cc im l hnh chiu ca im ( )4;3;2M trn cc trc ta .

Bi 2: TRONG MT PHNG OXYZ, Vit phng trnh mt phng i qua 3 im:a, A(2; -1; 3); B(0; 3; 1); C(-2; 4; 0) s: 3x + 5y - z - 14 = 0

b, A(2; 0; 0) ; B(0; 1; 0) ; C(1; 2; 3) s: x + 2y - z - 2 = 0

Bi 3: Vit phng trnh mt phng i qua im M(1; 1; 2) v cha ng thng d:

1 2

3

4 5

x t

y t

z t

= + = = +

Bi 4: Vit phng trnh mt phng i qua im M(1; -3; 2) v cha trc Oz

Bi 5: Vit phng trnh mt phng cha d1 v song song vi d2 bit

1 2

7 4 4 1 9 12: ; :

3 2 3 1 2 1

x y z x y z d d

+ + += = = =

Bi 6: Cho t din ABCD,bit A(0;1;2); B(1; 0; 5) ; C(3; -1; 0) ; D(5; -2; 1).a) Vit phng trnh mt phng i qua A v song song vi mt phng (BCD)

b) Vit phng trnh mt phng cha AB v song song vi CDc) Vit phng trnh mt phng cha AB v vung gc vi (P): 3x - y + 2z - 4 = 0

Bi 7: Vit PTMP tip xc vi mt cu (S): (x -1)2 + (y - 2)2 + (z + 3)2 = 16 v i qua giaotuyn ca hai mt phng (P): 3x + y - 2z + 1 = 0 v (Q): x - 2y + z - 3 = 0.Bi 8: Vit PTMP tip xc vi (S): (x -1)2 + (y - 2)2 + (z + 1)2 = 25 v vung gc vi ng

thng c phng trnh l:1 2

2 3 1

x y z + = =

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Tit 5-6 - TUN 5PHNG TRNH NG THNG

I, MC CH - YU CU:- Nm c dng ca phng trnh ng thng; xc nh c yu t trong phng

trnh ng thng. - Vit c phng trnh ng thng c bn.

II, CHUN B: H thng l thuyt; bi tp vn dng; bi tp rn luyn.

III, NI DUNG BI DY:

PHNG PHP NI DUNGBi 1. Tm vct ch phng v mt im thuc ng thng dtrong cc trng hp sau:

a,

1 2

: 3

x t

d y t

z t

= + =

=b,

2 1:

1 2 3

x y z d

+= =

c, d vung gc vi mp(P): 2x - 3y + z - 4 = 0 ti im M(2; -3; -5

Bi 2. Vit phng trnh ng thng:a, i qua hai im A(1; 2; -3) v B(0; 2; -1)

b, i qua hai im O v M(1; -1; 3)c, i qua im M(1; -1; -4) v song song vi ng thng

2 1:

1 2 3

x y z d

+= =

d, i qua im N(1; 0; 1) v vung gc vi mp (P):3x - 2y + z - 1= 0e, i qua gc to v vung gc vi mp: 2x - z = 0

Bi 3. Vit phng trnh ng thng i qua im A(2; 1; -3) vsong song vi hai mt phng(P):x+y-z-3=0; (Q): 2x - y + 3z + 1 = 0

Bi 4: Vit phng trnh ng thng i qua im A(1;1;3) v

song song vi mp(P):3x - 2y + z = 0 v mt phng Oxz.

* Cng c:+ Phng trnh tham s v phng trnh chnh tc ca ng thng.+ Cch vit mt s dng ton v PT ng thng.* Dn d:V gii li cc bi tp trn v hc k l thuyt c bn v gii thm mt s bi tp sau:

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BI TP RN LUYN VIT PHNG TRNH NG THNGBi 1: Cho 3 im A(1; 2; -1) ; B(2; 0; -2); C(2; 2; 0)

a, Vit phng trnh ng thng AB, BC, CA.b, Vit phng trnh ng thng i qua A v song song vi BC.c, Vit phng trnh ng thng vung gc vi mt phng (ABC) ti A.d, Vit phng trnh ng thng i qua O v vung gc mt phng (OAB)e, Vit phng trnh ng thng i qua M(2; 0; 1) v vung gc vi (ABC). Tm giao im ca chng.

Bi 2: Cho im A(1; -2; -1) v mt phng (P): 3x - 6y + 2z - 3 = 0a, Vit phng trnh ng thng i qua A v vung gc vi mp(P).

b, Vit phng trnh mt phng i qua A v song song vi (P).c, Vit phng trnh mt cu tm A v tip xc vi (P).

Bi 3: Cho im M(2; 2; -1) v ng thng d:3 1

2 1 2

x y z += =

a, Vit phng trnh ng thng i qua im M v song song vi d.b, Vit phng trnh ng thng i qua O v vung gc vi mt phng cha M v d.

Bi 4: Cho hai im ( ) ( )5;1;21;1;1 NM,

trong khng giana, Vit phng trnh ng thng MN.b, Tm giao im ca MN vi mt cu (S): 0642222 =++ zyxzyx

Bi 5: Trong khng gian vi h to O xyz cho 4 im A(3;2;6),B(3; -1, 0), C(0,-7,0), D(-2,1; -1). a/ Vit phong trnh mt phng (ABC). b/ Tnh gc gia ng thng (d) i qua hai im A, D v mp(ABC)

Bi 6: Trong khng gian, Cho :)(; 01 0

25

21

2

: =++

+=

+=

=

z -yx

tzty

tx

a, Vit phng trnh ng thng i qua I(1; 0; -1) v song song vi .b, Vit phng trnh ng thng i qua J(2; 0; -3) v vung gc vi mp( )c, Tm to giao im ca v mp( ).

---------------------

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Tit 7-8, Tun 5: PHNG TRNH MT CU

I, Mc ch - Yu cu:- Hc sinh nm vng cc tnh cht ca mt cu; dng phng trnh mt cu; cch x tm

v bk mC- Vit c phng trnh mt cu c bn.

II, Chun b: H thng l thuyt; bi tp vn dng; bi tp rn luyn.

III, NI DUNG BI DY:

PHNG PHP NI DUNGBi 1. Xc nh tm v tnh bn knh mt cu:1) 2 2 2 4 6 5 0 x y z x y+ + + =2) 2 2 2 8 2 1 0 x y z x z + + + + =S: 1)Tm I(2 ;-3 ;0) v R=3 2 2) Tm I (4;0;-1) v R=4

Bi 2. Lp phng trnh mt cu:1) Tm I(2;2;-3) v R=32) ng knh AB vi A(1;-3;5); B(-3; 4; -3)3) Tm A(1; 2; -1) v i qua im B(2; 0; 3)4) Qua A(3;1;0); B(5;5;0) v tm I thuc Ox

5) Tm I(2; -1; 1) v tip xc mt phng 2x - 2y + z - 1 = 0.6) Tm A(2; 2 ; 3) v tip xc vi mp Oxy.

Bi 3. Trong khng gian Oxyz, Cho: A(1; 2; -4)

d: 2 2 21

; ( ) : 2x 2 4z 3 0

2 1 3

x y z S x y z y

+= = + + + =

a, Vit phng trnh ng thng i qua A v song song vi d.b, Tm giao im ca ng thng d v mt cu (S).

* Cng c:+ Cch xc nh tm v bn knh ca mt cu;+ Cch vit phng trnh mt cu.* Dn d:V gii li cc bi tp trn v hc k l thuyt c bn v gii thm mt s bi tp sau:

-----------------

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Tit 9-10 - TUN 5V TR TNG I - GIAO IM

I. Mc ch - Yu cu:- Hc sinh bit xt v tr tng i gia hai mt phng; gia hai ng thng; gia t v

mp.- Bit tm giao im gia t vi mp; t vi mt cu.

II. Chun b:H thng l thuyt; bi tp vn dng; bi tp rn luyn.

III, NI DUNG BI DY:

PHNG PHP NI DUNGBi 1. Cho hai mt phng

( )( ) 04732:

052:

2

1

=+=++

zyx

zyx

a, Vit phng trnh ng thng i qua M(1; 1; 2) v vung gc vi 1

b, Chng minh rng hai mt phng trn ct nhau. Vit phng trnh ngthng giao tuyn ca hai mt phng .

Bi 2. Xc nh m, l 0742;0322 =++=+++ zymxzlyx song song vi nhau.

Hai mt phng song song vi nhau khi v ch khi7

3

4

2

2

2

== lm

=

=

=

=

1

4

4

2

2

4

22

l

m

l

m

Bi 3. Cho hai mt phng (P): 3x - 2y + z - 4 = 0; (Q): 2x + 3y+ 2 = 0a, Chng minh hai mt phng (P) v (Q) vung gc nhau.

b, Vit phng trnh ng thng i qua gc to v vunggc (P).c, Vit phng trnh giao tuyn ca hai mt phng trn.

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Bi 4. Cho hai ng thng 1 2

x 2tx 1 y z

( ) : y 1 t , ( ) :1 1 1

z t

= = = = =

V

v mt cu (S): 2 2 2x y z 2x 2y 4z 3 0+ + + + =a, Chng minh hai ng thng trn cho nhau.

b, Xc nh tm v tnh di bn knh ca mt cu (S).c, Vit pt tip din ca mt cu (S), bit tip din song songvi hai thng 1 2( )va( )

* Cng c:+ Phng php xt v tr tng i gia cc yu t: t, mp, mt cu trong khng gian.+ Cch tm giao im ca t vi mt phng, mt cu.

* Dn d:V gii li cc bi tp trn v hc k l thuyt c bn v gii thm mt s bi tp sau:

BI TP RN LUYNBi 1: Trong khng gian vi h ta Oxyz , cho im M(1;0;5) v hai mt phng (P) :

2x y 3z 1 0 + + = v (Q) : x y z 5 0+ + = .a. Tnh khong cch t M n mt phng (Q) .b. Vit phng trnh mt phng ( R ) i qua giao tuyn (d) ca (P) v (Q) ng thi vung gc vi mp(T) :3x y 1 0 + = .

Bi 2: Trong khng gian vi h ta Oxyz , cho thng (d ) :x 3 y 1 z 3

2 1 1

+ + = = v mp(P) :

x 2y z 5 0+ + = .a. Tm ta giao im ca ng thng (d) v mt phng (P) .b. Tnh gc gia ng thng (d) v mt phng (P) .c. Vit phng trnh ng thng ( ) l hnh chiu ca ng thng (d) ln mt phng (P).Bi 3: Trong khng gian vi h ta Oxyz , cho 4 im A( 2;1; 1) ,B(0;2; 1) ,C(0;3;0) , D(1;0;1) .a. Vit phng trnh ng thng BC .b. Chng minh rng 4 im A,B,C,D khng ng phng .c. Tnh th tch t din ABCD .Bi 4: Trong khng gian vi h ta Oxyz cho im M(1; 1;1) , hai ng thng

x 1 y z( ):1 1 1 4

= =

,

x 2 t

( ): y 4 2t2z 1

= = + =

v mt phng (P) : y 2z 0+ =

a. Tm im N l hnh chiu vung gc ca im M ln ng thng ( 2 ) .b. Vit phng trnh ng thng ct c hai ng thng ( ) ,( )1 2 v nm trong mt phng (P) .

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Bi 5: Trong khng gian vi h ta Oxyz , cho hai ng thngx 1 y 2 z

( ):1 2 2 1

= =

,

x 2t

( ): y 5 3t2

z 4

= = +

=

a. Chng minh rng ng thng ( )1 v ng thng ( )2 cho nhau .

b. Vit phng trnh mt phng ( P ) cha ng thng ( )1 v song song vi ng thng ( )2 .Bi 6: Trong khng gian vi h ta Oxyz cho im M(2;3;0) , mt phng (P ) : x y 2z 1 0+ + + = v mt cu

(S) : 2 2 2x y z 2x 4y 6z 8 0+ + + + = .a. Tm im N l hnh chiu ca im M ln mt phng (P) .b. Vit phng trnh mt phng (Q) song song vi (P) v tip xc vi mt cu (S) .

Bi 7: Trong khng gian vi h ta Oxyz , cho hai ng thng

x 2 2t

(d ) : y 31z t

=

= =v

x 2 y 1 z(d ) :2

1 1 2

= =

.

a. Chng minh rng hai ng thng (d ),(d )1 2 vung gc nhau nhng khng ct nhau .

b. Vit phng trnh ng thng i qua O v vung gc vi c hai ng thng (d ),(d )1 2 .

Bi 8: Cho mt phng ( ) : 2x y 2z 3 0 + = v (d1 ) :x 4 y 1 z

2 2 1

= =

, (d2 ) :

x 3 y 5 z 72 3 2+ + = =

a. Chng t ng thng ( d1) song song mt phng ( ) v ( d2) ct mt phng ( ) .

b. Tnh khong cch gia ng thng ( d1) v ( d2 ).

c. Vit phng trnh t ( ) song song vi mp( ) , ct ng thng ( d1) v ( d2 ) ln lt ti M v N sao cho MN=3.

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Tun 6, Tit 1- 6 KHO ST HM S - BI TON LIN QUANI, Mc ch yu cu:Rn luyn k nng kho st hm s v gii cc bi ton lin quan n kshs.

II, Ni dung n tp:

1. Bin php thc hin:- Rn luyn k nng kshs /v cc hm s c cha phn s, nghim cha cn thc.- Rn luyn k nng vit pttt v bin lun s nghim, bi ton v s tng giao.2. Bi tp:

Bi 1: Cho hm s 3 21 3 54 2

y x x = - +


b) Vit PTTT ca th (C) ti im c honh bng -2.

c) Vit PTTT ca th (C) bit rng tip tuyn song song vi ng thng

93

4y x= - +

d) nh m phng trnh 3 21 3 5 2 04 2

x x m- + - = c ng 1 nghim

Bi 2: Cho hm s 3 23 y x x = - +

a) Kho st v v th (C) ca hm s.b) Vit PTTT ca th (C) ti im c tung bng 4.

c) Tnh din tch hnh phng c gii hn bi (C ) , trc 0x v x=1Bi 3: Cho hm s 4 24 2 y x x = - +


b) Vit PTTT ca th (C) ti im c honh bng 3 .

Bi 4: Cho hm s 4 21 2 44

y x x = - + .



Bi 5 : Cho hm s y =3

2 1

x

x

+

.


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b) Lp pttt ca th (C) bit rng tip tuyn song song vi ng thng

y=x.

Bi 6 : Cho hm s y = 2 4

1

x

x

+

-


b) Lp pttt ca th (C) ti im c tung bng 1.

3, Bi tp t rn luyn:Bi 1. Cho hm s y = - x3 + x 1

a, Kho st v v th hm s trn.b, Vit phng trnh tip tuyn ti giao im ca (C) vi trc tung.

Bi 2. Cho hm s y = -2x4 + 4x2 + 1a, Kho st v v th (C) ca hm s trn.

b, Tnh din tch hnh phng gii hn bi (C) v trc honh.

Bi 3. Cho hm s1

2

xy

x

=+

a, Kho st v v th (C) ca hm s trn.b, Vit phng trnh tip tuyn ca (C) ti im c tung bng 3.c, Tnh din tch hnh phng gii hn bi (C) trc tung, trc honh.

4, Rt kinh nghim:BI KIM TRA TUN 6

Cu I (3,0 im): Cho hm s:

42

42

x

y x= - -

1) Kho st s bin thin v v th ( )C ca hm s.2) Tnh din tch hnh phng gii hn bi th ( )C v trc honh.3) Tm m phng trnh sau y c ng 2 nghim phn bit: 4 22 2 0 x x m- - =

Cu II (3,0 im):

1) Tm nguyn hm ( )F x ca 21

( ) 3 4 xf x x ex

= - + bit rng (1) 4F e=

2) Tnh tch phn:

a)

3ln2

0 1

x

xeI dxe+= b)

33

0 2 1

xdx

x=

+Cu III (1,0 im): Tnh din tch hnh phng gii hn bi cc ng sau y:

2 2( 1) , y x x y x x = - = + v 1x=-Cu IV (2,0 im): Trong khng gian Oxyz, cho 4 im

( 1;1;1), (5;1; 1), (2;5;2), (0; 3;1) A B C D- - -

1) Vit phng trnh mt phng (ABC). T chng minhABCD l mt t din.

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2) Vit phng trnh mt cu (S) c tm l imD, ng thi tip xc vi mt phng(ABC). Vit phng trnh tip din vi mt cu (S) song song vi mp(ABC)

Cu V (1,0 im): Tnh 1 2x x+ , bit 1 2,x x l hai nghim phc ca 23 2 3 2 0x x- + =Tun 7, Tit 1, 2 GI TR NH NHT GI TR LN NHT CA HM S

I, Mc ch yu cu:

- Hc sinh bit cc bc xc nh GTNN, GTLN ca hm s trn on cho trc.- Bit c phng php tm GTNN, GTLN trn khong.

II, Ni dung n tp1. Bin php thc hin

- Gio vin nhc li pp gii bi ton tm GTLN, GTNN trn on, khong.- Hc sinh nhn dng c bi ton tm GTLN, GTNN.

2. Bi tp:LOI 1 : TRN KHONG

1) Tm gi tr ln nht gi tr nh nht ca cc hm s sau:y = 4 - x2; b) y = 4x3 3x4; c) y = x4 + 2x2 2;

d) y = 2xx2 ++ ; e) y =x

1xx2 ++vi x > 0; g) y =

2x +3x 1

x-1

+vi x < 1.

h)1

y =cosx

trn khong ;3

2 2

k) = 2x

yx +4

Tm kch thc ca hnh ch nht c din tch ln nht, bit rng chu vi bng 16 cm.HD: - Gi mt kch thc l x, iu kin 0 < x < 8

Din tch ca hnh ch nht l S(x) = x( 8 x).- Tm x(0; 8) S(x) ln nht. S: x = 4 cm

2) Hy xc nh hnh ch nht c chu vi nh nht, bit din tch bng 48cm2.HD: - Gi x l mt kch thc ca hnh ch nht, iu kin x > 0.

- Chu vi ca hnh ch nht l 48( ) 2( )P x xx

= + .

- Tm x(0; + ) P(x) nh nht. S: Hnh vung c cnh bng 4 3mLOI 2 : TRN ON

1) y = x4 - 4x2 + 5 trn [1; 2] 2) y = x3 - 6x2 + 9x trn [2; 4]3) y = x + 22 x 4) y = 1 1x x + +5) y = 2 2x 3xe + trn [0; 2] 6) y = e2x + ex - 3x trn [-1; 1]

7) 2f(x) x 4x= - + trn [ 2;3]- . 8) y = cos2x cosx + 2

9) f(x) = 2 sinx + sin2x trn on 30; 2

10) 3 2y 2sin x cos x 4sinx 1= + +

11) 2 cos y x x= + trn on [0; ]2

. 12) 2( 2). 4= + y x x

13) y = lnx x . 14) 2x 1

y

1 x

+=

+

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15)xe

yxe e

=+

trn on [ln2;ln4]

III, Rt kinh nghim tit dy:Tun 7, Tit 3 4 PHNG TRNH, BPT M LGARIT

I, Mc ch yu cu:- Hc sinh nm vng cng thc, bit vn dng vo bi ton c th.

- p dng v nhn dng tt cc cng thc vo gii pt, bpt m, lgarit.

II, Ni dung:

1, Bin php thc hin:

- a ra bi tp tng hp hc sinh nhn dng.

- Nhn mnh li thng gp, phn tch tm hng gii cc bi ton.

2, Bi tp:

Bi 1.Gii cc phng trnh sau:2 1) 4 8x xa -= S:

3

4x=

24 2 5 3) 3 9 x x x b - - -= S: 1; 3x x= = -

2 2 3 3) 2 .5 2 .5 x x x x c + + = S: 1x=5) 3 4xd - = S: 35 log 4x= +

e)3 2 1

7 90

9 7

x x + =

f) 3x+1 - 2.3x 4.3x 1 = - 3

Bi 2. Gii cc phng trnh sau:) 4 3.2 10 0x xa + - = S: 1x= ) 25 23.5 5 0

x xb - - = S: 2x=( )2 1) 3 82.3 9 0x xc + - + = S: 2x=

2 2) 3 3 30x xd + -+ = S: 1x= 1 1) 5 5 26x xe + -+ = S: 1x= ( ) ( )) 2 3 2 3 14

x x

f - + + = S: 2x=

Bi 3. Gii cc phng trnh sau:

3 9) 3log log 5a x x- = S: 9x= 3 9 275

) log log 3 log3

b x x x + + = S:7

113x=

( )5 5) log 1 log

1

xc x

x

- =

+S:

1 5

2x

+= ( )2 8

1) log 5 2log 3 1

3d x x- + - = (X =1)

Bi 4. Gii cc phng trnh sau:2) 9lg 10lg 1 0a x x- + = 2

5 5

1 3) log log 0

2 2b x x+ - =

3

10) log log 3

3xc x+ =

22 2) 2log 3log 11 0

4

xd x- - =

4, Rt kinh nghim tit dy:

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TUN 7 S PHCTIT 5,6I, Mc ch yu cu:

- Hc sinh nm vng cc khi nim lin quan n s phc v php tnh trn tp s phc.- Bit gii phng trnh bc hai trn tp s phc.- Tm cc s phc z tho mn yu cu cho trc.II, Bi tp:Bi 1. Cho s phc z1 = 2 3i; z2 = 3 + i

a, Tm mun ca z1.z2.

b, Xc nh phn thc v phn o ca s phc z = 12

2

3

z

z

+

c, Tm nghch o ca hai s phc trn.

Bi 2. Gii phng trnh trn tp s phc:a, x2 2x + 2 = 0 b, 2x2 5x + 4 = 0c, x2 4x + 7 = 0 d, -2z2 + 4z 3 = 0

Bi 3. Tm x, y sao cho: 1, (2x 3) + (3y + 1)i = 4 + 5i 2, (2x y) + (2x + 3i)i = 3 2iBi 4. Tm s phc z sao cho:

a, iz 432 += b, z3 = i

c, mun bng 52 ; t s gia phn phn thc v phn o bng21 .

d, (1+i)z + (2 - i)(1+3i) = 2+3i

Bi 5. Tm 2 s phc bit tng ca chng bng 2 v tch ca chng bng 3.

III, PHNG PHP THC HIN:- Khi qut li kin thc l thuyt trng tm;- Gi cc hc sinh yu thc hin cc bi ton n gin nh: php tnh trn s phc, xc nh

phn thc, phn o v mun ca s phc.

- Phn tch cc dng ton v nhng sai lm thng gp trong gii ton s phc.- Cho hc sinh lm cc bi tp tng t ti nh, gio vin kim tra bi lm hs.IV, BI TP RN LUYN:Bi 1. Cho s phc z1 = 3 3i; z2 = 2 + i

a, Tm mun ca z1 -2z2.

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b, Xc nh phn thc v phn o ca s phc z = 1 22 2

2

2

z z

z z

+

c, Tm nghch o ca hai s phc trn.

Bi 2. Gii phng trnh trn tp s phc:a, 2x2 3x + 5 = 0 b, -2x2 x 4 = 0c, z2 4z + 8 = 0 d, -2z2 + 4z 8 = 0

Bi 3. Tm x, y sao cho:1, (2x 3) + (3y + 1)i = x + 4 + 5yi 2, (2x + y) + (2x - y)i = 3y 2i

Bi 4. Tm s phc z sao cho:a, 2 3 4z i= b, (x2 2x + 3)(x2 + 2) = 0 c, x3 2x2 + 3x 2 = 0c, 2x4 6x2 8 = 0 d, (1+i)z + (2 - i)(1+3i) = 2+3i e, 2z - |z| = 1 + 8i

V, NH GI RT KINH NGHIM TIT N TP:

BI KIM TRA KIN THC TUN 2CU 1: (3 )

Cho hm s y =2x 1

1 x

(C)

a) Kho st v v th (C) ca hm s.b) Vit phng trnh tip tuyn ca (C) ti im c tung bng -1.

CU 2: (2 )

1) Tnh tch phn: I = ( )2

1

11 3ln x . x

e

dx+

2) Gii phng trnh trn tp s phc: 3iz + 2 4i = z (1+i)CU 3: (3 )

Trong h trc Oxyz, Cho ng thng d:1 2

2 1 3

x y z += = v mp(P): 3x 3y z + 19 = 0

a) Vit phng trnh ng thng i qua M(1; -2; 1) v vung gc vi (P).b) Vit phng trnh mt phng cha d v vung gc vi (P).c) Chng minh ng thng d v mp(P) song song nhau.

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TUN 8 TIT 1 6 NGUYN HM TCH PHN - NG DNGI, Mc ch yu cu:- Nhn dng c phng php tnh tch phn.- Thuc v p dng c cc cng thc tnh nguyn hm, o hm, vi phn ca hm s.- Rn luyn k nng tnh ton.- Nm vng phng php gii bi ton ng dng ca tch phn.II, Ni dung:1, Bin php thc hin:- Nhn mnh kin thc l thuyt lin quan;- H thng cng thc p dng;- Bi tp a dng, tng t.2, Ni dung n tp:

Dng 1: PHNG PHP PHN TCH S DNG NGUYN HM C BN.2 3 2

21

2 2 1 x x x

x

+ + ; ( )

20

1

2

1

x

x

+ ;

1

2

0

24 5

xdx

x x

;3

2 2

6sin cos

dx

x x

;2

0

sin2 .cos5 x xdx

2

1 1 1

dx

x x+ + ;34

2

0

1 cos

cos

xdx

x

;

22

0

sin xdx

;4

2

0

tg xdx

; ( )1

2009

0

1 x x dx .

Dng 2: PHNG PHP I BIN S DNG IBi tp:

1

22

0

1 x dx

;

1

20 1

dx

x+ ;

22

1

4 x dx

;

2 2

2

20 1

xdx

x; ( )

1 32

0

1 x dx

;

( )

222

302

1

x dx

x

2

2 2

04 x x dx ;

32

1 22 1

dx

x x ;

21

20 4

x dx

x ;

3

20 3

dx

x +

Dng 3: PHNG PHP I BIN DNG II3

0sin cos x xdx

; 320

sin xdx

; 320 cos xdx

; 20sin

1 cos

xdx

x

+ ;2

4

0

1 2sin

1 sin2

xdx

x

+

1

3 2

01 x x dx ;

15 3

01 x x dx ;

37

20 1

x dx

x+ ;

2 3

25 4

dx

x x + ;

31

20 1

x dx

x +

( )

ln3

30

1

x

x

e dx

e + ; ( )1 65 3

01 x x dx ;

1

0 2 1

xdx

x +

Dng 4: PHNG PHP TNH TCH PHN TNG PHN

( )20

1 sin I x x

= + ; ( )240 2cos 1 I x x

= ; ( )1

2

01 x I x e dx= ;

2

21

ln x I dx

x=

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( )3

2

2ln I x x dx= ; 34

0sin4x I e xdx

= ; ( )1

2

02 x I x x e dx

= + ; ( )1

2 2

04 2 1 x I x x e dx=

20

sin I x xdx

= ; 2 21 lne

I x xdx= .

NG DNG TCH PHN TNH DIN TCH HNH PHNG

Dng 1:

1) Tnh ?DS = , bit D gii hn bi th: 2 2 y x x= , 1, 2x x= = v trc Ox .

2) Tnh ?DS = , bit { }, 0, 1, 2x D y xe y x x= = = = =

3) Tnh ?DS = vi { }2 4 , 1, 3D y x x x x= = = =

4) Tnh ?DS = , vi , 0, , 03

D y tgx x x y = = = = =

5) Tnh ?DS = , 2ln

, 0, 1, 2x

D y y x xx

= = = = =

6) Tnh ?DS = ,ln

1, , 0,2

xD x x e y y

x

= = = = =

7) Tnh ?DS = 2 3 1

, 0, 1, 01

x x D y x x y

x

+ += = = = = +

8) Tnh ?DS = ,2 3sin cos , 0, 0,

2 D y x x y x x

= = = = =

Dng 2:

1) Tnh ?DS = , ( ){ }5

1 , , 0, 1x D y x y e x x= = + = = =

2) Tnh ?DS = , 2 21 1

, , ,sin cos 6 3

D y y x xx x

= = = = =

3) Tnh ?DS = , [ ]{ }22 sin , 1 cos , 0; D y x y x x = = + = +

4) Tm b sao cho din tch hnh phng gii hn bi th ( )2

2:

1

xC y

x=

+v cc ng

thng 1, 0, y x x b= = = bng4

Dng 3:1) Tnh ?HS = , vi { }, , 1

x xH y e y e x= = = =

2) Tnh ?HS = , { }21 , , 1 H y x x Ox x= = + =

3) Tnh ?DS =3 1

, ,1

xD y Ox Oy

x

= =

4) Tnh din tch hnh phng gii hn bi : 2 ; 3 ; 0x y y x x= = =

5) Tnh ?HS = , { }, 2 0, 0 H x y x y y= = + = =

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Dng 4:Tnh din tch hnh phng ( )D gii hn bi: ( ) ( );y f x y g x= =

1) Tnh din tch hnh phng gii hn bi: 2 2 y x x= ; 2 4 y x x= +2) Tnh din tch hnh phng gii hn bi: 2 2 y x x= + v 3y x=

3) Tnh din tch hnh phng gii hn bi:2

2 0 y y x + = v 0x y+ =4) Tnh din tch hnh phng gii hn bi: 2 5 0y x+ = v 3 0x y+ =

5) Tnh din tch hnh phng gii hn bi: 2 4 3 y x x= + v 3y x= +

6) Tnh din tch hnh phng gii hn bi2

44

xy = v

2

4 2

xy =

NG DNG TCH PHN TNH TH TCH

Dng 1:

1) Cho hnh phng D gii hn bi : , 0, 0, 3 D y tgx y x x

= = = = = a) Tnh din tch hnh phng D

b) Tnh th tch vt th trn xoay sinh ra khi D quay quanh trc Ox2) Tnh th tch ca vt th trn xoay sinh ra bi php quay xung quanh Oy ca hnh

gii hn bi Parabol ( )2

: ; 2; 42

x P y y y= = = v trc Oy

3) Cho hnh phng ( )D gii hn bi ( ) 2: 8 P y x= v ng thng 2x = . Tnh th tch

khi trn xoay khi ln lt quay hnh phng ( )D quanh trc Ox v trc Oy .Dng 2:

1) Tnh OxV bit: { }ln , 0, 1,D y x x y x x e= = = = =

2) Cho D l min gii hn bi th 2 ; 0; 0;4

y tg x y x x

= = = =

a) Tnh din tch min phng Db) Cho D quay quanh Ox , tnh th tch vt th trn xoay c to thnh.

3) Tnh OxV bit:3

2,3

x D y y x

= = =

4) Tnh OxV bit:4 40; 1 sin cos ; 0,

2 D y y x x x x

= = = + + = =

5) Tnh OxV bit: { }2 5 0; 3 0 D x y x y= + = + =

6) Tnh OxV bit: { }22 ; 2 4 D y x y x= = = +

7) Tnh OxV bit: { }2 24 6; 2 6 D y x x y x x= = + = +

8) Tnh OxV bit: { }2; D y x y x= = =

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BI KIM TRA TUN 3

1) Cho hm s2 3

1

xy

x

+=

+

a, Kho st v v th (C) ca hm sb, Vit phng trnh tip tuyn ca th hm s

2 3

1

xy

x

+=

+, bit tip tuyn song

song vi ng thng y x= .2) Gii phng trnh:

( )4 2/ 2.16 17.4 8 0; / log 2 log ;x xa b x x + = + =

3) Tnh tch phn:

I =3

7

20 1

x dx

x+ J = ( )

12

02 1 x x e dx

4) Cho hnh chp S.ABCD c y ABCD l hnh vung cnh a, cnh bn SA vung gcvi y, cnh bn SB bng a 3 . Tnh th tch khi chp S.ABCD theo a .

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Tun 4, Tit 1-2 HNH HC KHNG GIAN

I, Mc ch yu cu:

- Hc sinh nm vng cc bc gii bi ton hnh hc khng gian, bit cch v hnh biu dintrong khng gian, phn tch bi ton.

- Thuc c cng thc tnh din tch, th tch v cc cng thc lin quan.

- Nm c cch xc nh gc trong khng gian.

- Kh nng phn tch, tng hp v s dng kin thc hp l.

II, Ni dung:

1, Bin php thc hin:

- Nhc li s lc cch v hnh, cc cng thc lin quan.

- a ra h thng bi tp c bn.

- Vn dng v rn luyn ti nh.

2, Bi tp:

Bi 1. Cho hnh chp u S.ABC c cnh y bng a v cnh bn bng 2a. Tnh th tch ca

khi chp u S.ABC.

Bi 2. Cho hnh chp u S.ABC c cnh y bng a. Gc hp bi cnh bn v mt y bng

60 . Tnh th tch ca khi chp u S.ABC.

Bi 3. Tnh th tch hnh chp t gic u c y l hnh vung cnh bng a v cc cnh bn

u bng nhau v bng 2a .

Bi 4. Cho hnh chp SABC c ABC vung ti B, AB = a, BC = 2a. ( )SA ABC v

2SB a= . Tnh th tch ca khi chp SABC.Bi 5. Cho hnh chp SABC c ABC l tam gic u cnh a. ( )SA ABC . Bit 5SC a= . Tnh

th tch ca khi chp SABC.

Bi 6. Cho hnh chp SABC c ABC vung ti B, AB = 3, BC = 4. ( )SA ABC v SC hp

vi (ABC) mt gc bng 30 . Tnh th tch ca khi chp SABC.

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Bi 7. Cho hnh chp u S.ABCD c cnh y bng a. Gc hp bi cnh bn v mt y bng

45 . Tnh th tch ca khi chp u S.ABCD.

Bi 8. Cho hnh chp S.ABCD c y ABCD l hnh vung c cnh a v SA vung gc y.Mt bn ABCD v mt bn (SCD) hp vi y mt gc 600.

a) Tnh th tch hnh chp S.ABCD.b) Tnh khong cch t A n mt phng (SCD).

Bi 9. y ABCD ca hnh chp S.ABCD l mt hnh vung cnh a. Hai mt bn (SAB) v(SAD) cng vung gc vi mt y. Gc gia SC v (SAB) bng 300.

a) Tnh th tch khi chp S.ABCD.b) Xc nh tm v tnh din tch mt cu ngoi tip hnh chp S.ABCD.

Bi 10. Cho hnh chp tam gic S.ABC c y ABC l tam gic vung ti nh B, cnhbn SA vung gc vi y. Bit SA = AB = BC = a. Tnh th tch ca khi chp S.ABCtheo a.3, Bi tp rn luyn:

Bi 1. Cho t din ABCD c y ABC l tam gic u cnh a, cnh bn AB vung gc vi yc di bng 2a. Tnh th tch khi t din ABCD.Bi 2. Cho t din ABCD c y ABC l tam gic u cnh a, cc cnh bn bng nhau v bng2a. Tnh th tch khi t din ABCD.Bi 3. Cho khi chp S.ABC c ABC l tam gic u cnh 2a, cnh bn SA vung gc vi y.Tnh th tch khi chp bit gc gia cnh bn SC vi mt y bng 60o.Bi 4. Cho khi chp S.ABC c ABC l tam gic u cnh a, cnh bn SA vung gc vi mty. Tnh th tch khi chp bit gc gia mt bn SBC v mt y bng 450.

Bi 5. Cho hnh chp S.ABC c y ABC l tam gic vung ti A, cnh bn SA vung gc vimt y. Tnh th tch khi chp bit SA = AB = a; BC = 3a.Bi 6. Cho hnh chp S.ABC c y l tam gic vung cn ti A, cnh bn SA vung gc vimt y v bng a 2 . Gc gia cnh bn SB v mt y bng 300. Tnh th tch khi chp.Bi 7. Tnh th tch t din ABCD c BCD l tam gic vung cn ti B, bit cnh BC bng 2av cnh bn AB vung gc vi y; gc gia mt bn ACD v mt y u bng 60o.Bi 8. Cho hnh chp S.ABCD c y ABCD l hnh vung cnh a, cc cnh bn bng nhau v

bng 3a. Tnh th tch khi chp.Bi 9. Cho khi chp S.ABCD c y ABCD l hnh vung c di ng cho bng 2a,cnh bn SA vung gc vi y, bit cnh SB hp vi y mt gc 45 o. Tnh th tch khichp.Bi 10. Cho hnh chp S.ABCD c y ABCD l hnh vung cnh 2a, cnh SA vung gc viy, gc gia mt bn SBC v y bng 30o. Tnh th tch khi chp.Bi 11. Cho hnh chp S.ABCD c y ABCD l hnh vung cnh 2a, cnh SA vung gc viy, gc gia mt bn SBD v y bng 60o. Tnh th tch khi chp.4, Rt kinh nghim:

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Tun 4, Tit 3-4-5-6 PHNG PHP TO TRONG KHNG GIAN

I, Mc ch yu cu:

- Khi qut li ton b kin thc PP to trong KG.

- Rn luyn k nng phn tch, nhn dng tm cc yu t vit c phng trnh mp, mc, t.- Lm quen cc dng ton lin quan nh xt v tr tng i, tm giao im.

II, Ni dung:

1, Bin php n tp:

- Hc sinh t ti hin li cc kin thc l thuyt trn c s cc bi ton gv a ra;

- Rn luyn k nng ca hc sinh thng qua vic phn tch, tnh ton.

2, Bi tp:

Bi 1. Cho cc im A(1; 1; 2); B(3; -1; -2); C(0; 1; 2); D(4; 1; -2)

1, Tnh di cc cnh AB, AC, BC, CD.

2, Chng minh rng 4 im ABCD lp thnh t din.

3, Vit phng trnh mt cu tm A v tip xc mp(BCD)

4, Vit phng trnh ng thng AB, CD.

5, Tm to chn ng cao ca t din h t nh A.

6, Vit phng trnh ng thng ni trung im ca AB, CD.

Bi 2. Trong khng gian cho mt cu (S): 05624222 =+++++ zyxzyx v ng

thng d:

2

1 2

3 2

x t

y t

z t

= + = = +

a, Xc nh tm I v bn knh R ca mt cu (S).

b, Vit phng trnh mt phng i qua I v vung gc vi ng thng d.

c, Tm giao im M, N ca d v mt cu (S). Tnh di MN.

Bi 3. Vit ptmp i qua A(3;1;-1) ; B(2;-1;4) v vung gc vi mp(P): 2x y + 3z 1 = 0

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Bi 4. Trong h ta Oxyz , cho hai ng thng

x 2 2t

(d ) : y 31z t

= = =

vx 2 y 1 z

(d ) :21 1 2

= =

a. Chng minh rng hai ng thng (d ),(d )1 2 vung gc nhau nhng khng ct nhau.b. Vit phng trnh mt phng cha ng thng d1 v song song d2.

c, Vit phng trnh ng thng vung gc vi c (d ),(d )1 2

4, Rt kinh nghim ni dung n tp

BI KIM TRA TUN 4

Cu 1. (3.0 im) Cho hm s y = - 2x3 + 3x2 1a) Kho st v v th (C) ca hm s.

b) Vit phng trnh tip tuyn ca (C) ti im c honh bng -2.

Cu 2. (1.0 im) Tnh tch phn: I =

Cu 3. (1.0im) Tm s phc z bit 3z 2i = 7 + 4iCu 4. (2.0 im) Trong h trc Oxyz, cho A(1; -1; 2) ; B(2; -1; 1); mt phng (P): 3x y + 2 = 0

a, Vit ph.trnh ng thng i qua hai im A v B. Tm to giao im caAB v (P).

b, Lp phng trnh mt phng cha AB v vung gc vi mp(P).****************

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