Đe va huong dan giai toan khong chuyen.pdf

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V Tin Trnh Trng PhThng Nng Khiu

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TON KHNG CHUYN TUYN SINH VO LP 10 TRNG PHTHNG NNG KHIU I HC QUC GIA TP.HCM T2010 N

2015 V HNG DN GII

Ton khng chuyn 2009 2010.

Bi 1.(2 im)

a)

Gii phng trnh bng cch t n s5

4

x

t x .

2

2

400 535 24

4

xx

xx

.

b) Cho phng trnh 2 3 1 2 3 0mx m x m .

Tm m phng trnh c hai nghim phn bit 1 2,x x tha mn2 21 2 34x x .

Bi 2.(2,5 im)

Xt biu thc2 3 3 4 5

1 5 4 5

x x x xR

x x x x

.

a) Rt gn R .b) Tm sthc x 2R . Tm stnhin x l schnh phng sao cho R

l snguyn.

Bi 3.(2 im)

a) Gii hphng trnh:2 2

0

8

x xy y

x y

b) Cho , ,a b c l di ba cnh ca tam gic ABC. Gisphng trnh

0x a x b x b x c x c x a c nghim kp. Tnh s

o cc gc ca tam gic.


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Bi 4.(1,5 im)

Cho tam gic ABC c 0 0

60 , 45ABC ACB . Dng AH BC H BC ,

v dng HK AB K AB . Gi M l trung im ca AC.

Bit 3AH , tnh BC v chng minh BKMC l tgic ni tip.

Bi 5.(1 im)

Trong kkim tra mn Ton, mt lp gm ba tA, B v C. im trung bnh cahc sinh cc tc thng k bng sau :

T A B C A v B B v C

im trung bnh 9.0 8.8 7.8 8.9 8.2

Bit tA gm 10 hc sinh, hy xc nh shc sinh v im trung bnh ca tonlp.

Bi 6.(1 im)

Cho tgic li ABCD ni tip ng trn (O), c nh A cnh v cc

nh B, C, D di chuyn trn (O) sao cho 090ABD . Ktia Ax vung gc vi AD

ct BC ti E, ktia Ay vung gc vi AB ct CD ti F. Gi K l im i xng caA qua EF. Chng minh tgic EFCK ni tip c v ng thng EF lun i quamt im cnh.

Hng dn gii.

Bi 1.

a)

iu kin: 0x

Vi2 2 2

2 2

2 2

5 5 25 5 25 5

4 4 16 2 16 2

x x x xt t t

x x x x

2 2

2

40016 40x t

x


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Thay vo phng trnh ta c: 2

1

416 24 5 0

5

4

t

t t

t

Vi1

4t ta c phng trnh : 2

55 120 0

44 4

xxx x

xx

Vi5

4t ta c phng trnh 2

5 105

5 5 25 20 0

4 4 5 105

2

xx

x xx

x

So vi iu kin, phng trnh 4 nghim5 105 5 105

; ; 5;42 2

b) Phng trnh c hai nghim phn bit

2 2

0 00

0 17 6 9 09 1 4 2 3 0

m mm

m mm m m

2

0

03 14417 0

17 17

m

mm

Theo inh l viet ta c:

1 2 1 2

3 1 3 2,

m mx x x x

m m

Ta c: 22 2

1 2 1 2 1 234 2 34x x x x x x

2 22

19 1 3 22 34 21 12 9 0 3

7

mm mm m

mm m

So vi iu kin, nhn3

1,7

m m


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Bi 2.

a) iu kin xc nh : 0x v 25x

Rt gn2

5

xR

x

x

b)12 02 12

2 2 0 05 5 5 0

xx xR

x x x

hoc

12 0

5 0

x

x

144x hoc 25x .

So vi iu kin xc nh ta c 144x hoc 0 25x .

Ta c

7

1 5R x

V x l schnh phng nn x l snguyn, do R l snguyn7

5x

l snguyn 5x l c ca 7 5 7; 1;1;7x

5 7 2x x (loi).

5 1 4 16x x x (nhn)

5 1 6 36x x x (nhn)

5 7 12 144x x x (nhn)

Vy 16;36;144x

Bi 3.

a)

t ,u x y v xy ta c 22 2 22 2x y x y xy u v

Gihphng trnh2 2

0 4

42 8 2 8 0

u v u v u

vu v v v

hoc

2

2

u

v

Vi 4, 4u v ta c ,x y l nghim ca 2 4 4 0 2X X X


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2

2

x

y

Vi 2, 2u v ta c ,x y l nghim ca 21 3

2 2 01 3

xX X

x

1 3

1 3

x

y

hoc

1 3

1 3

x

y

Vy hc nghim 2; 2 , 1 3;1 3 , 1 3;1 3

b)

Phng trnh cho bin i thnh 23 2 0x a b c x ab bc ca

2 2 2 2' 3a b c ab bc ca a b c ab bc ca

2 2 21

2a b b c c a

Phng trnh c nghim kp '' 0 a b c ABC u.

Bi 4.

+ Tnh BC.

Tam gic AHC vung cn ti H nn 3AH HC


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Tam gic ABH vung ti H c 00

360 1

tan60 3

AHABH BH

Do 1 3BC BH CH .

+ Chng minh BKMC ni tip.

V tam gic AHC vung cn ti H nn HM AC

Tgic AKHM ni tip ng trn ng knh AH

AKM AHM ACB tgic BKMC ni tip ng trn.

Bi 5.Gi shc sinh ca tB v C ln lt l *, ,x y x y

Ta c hphng trnh

10.9 .8,88,9

1010

.8,8 .7,8 158,2

x

xx

x y y

x y

Bi 6.


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+ Chng minh EFCK ni tip

Nhn xt : 090EAF EAB (Ay vung gc AB)

0

90EAF FAD (Ax vung gc AD)

Nn 0180EAF EAB EAF FAD hay 0180EAF DAB

Do EAF BCD

M EAF EKF (A i xng vi K qua EF)

Do EKF BCD EFCK ni tip.

+ Chng minh EF qua im cnh.

Ta c 0 090 , 90BAK KAF AFE KAF AFE BAK

Do tgic EFCK ni tip

BCK EFK AFE (do A, K i xng qua EF) BAK

tgic BACK ni tip suy ra K thuc ng trng (O)

Do AK l dy cung ca (O) nn EF l trung trc ca dy cung AK si quaim O cnh.


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Ton khng chuyn 2011 - 2012.

Bi 1(2,5 im). Cho phng trnh 2 2

2 3 0 1x mx m x

a) Gii phng trnh (1) khi 2m .

b) Tm m phng trnh 2 22 0x mx m c hai nghim phn bit 1 2,x x

tha 2 2 21 22 7 2x x m .

c) Chng minh phng trnh (1) lun c khng qu hai nghim phn bit.

Bi 2(2 im).

a)

Gii phng trnh 2 5 2 1 6x x x

b) Gii hphng trnh2 2 2 1

1

x y y

xy x

Bi 3(1,5 im)

a) Rt gn biu thc

23 11 1

:11 1

x xx xR

xx x

vi 0x v

1x .

b) Chng minh 1R .

Bi 4(1 im).

Mt tmua nguyn vt liu tchc thuyt trnh ti lp ht 72.000 ng, chi phc chia u cho mi thnh vin ca t. Nu tgim bt 2 ngi th mi ngi

phi ng thm 3.000 ng. Hi sngi ca t?

Bi 5(3 im).

Tam gic ABC c 0 075 , 45 , 2,BAC BCA AC a AK vung gc vi BC (K

thuc BC).

a) Tnh di cc on KC v AB theo a.


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b) Gi H l trc tm v O l tm ng trn ngoi tip tam gic ABC. TnhOHC.

c) ng trn tm I ni tip tam gic ABC. Tnh bn knh ng trn ngoitip tam gic HIO theo a.

Hng dn gii

Bi 1.

a) Khi 2m , phng trnh trthnh 2 2 8 3 0x x x

iu kin 3x .

Phng trnh cho tng ng vi2

332

2 8 04

xxx

x xx

So vi iu kin, ta c phng trnh c nghim l 3, 4x x

Vy phng trnh cho c nghim l 3, 4x x

b)

Phng trnh 2 2 2 22 0 2 2 0x mx m x mx mx m

2

2 0x m

x m x mx m

Phng trnh c hai nghim phn bit 0m

+ trng hp 1. 1 22 ,x m x m , ta c phng trnh 2 2 22 2 7 2m m m

2

2 0m

(v nghim m).

+ trng hp 2. 1 2, 2x m x m , ta c phng trnh 2 2 22 2 7 2m m m

2 11

1

mm

m

(tha mn iu kin)


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Vy gi trm cn tm l 1; 1m m

c) Phng trnh 2 23 0 3

1 2 0 2 ,

33 0

x x

x mx m x m x m

xx

V 22 2 0m m m nn hai s 2m v m chc mt skhng m.

Do phng trnh 2 22 0x mx m c ti a mt nghim khng nhhn 3. Do phng trnh (1) c khng qu hai nghim phn bit.

Bi 2.

a)

Phng trnh 2 5 2 1 6x x x .

iu kin5

22

x

Bnh phng hai vta c phng trnh tng ng

7 2 2 5 2 7 2 6x x x x x

2 5 2 6x x x

2 2 110 2 6 2 2 4 0

2

xx x x x x

x

(tha iu kin)

Vy phng trnh cho c hai nghim 1; 2x x

b) Gii hphng trnh2 2

2 1

1

x y y

xy x

T 1 2 2 1xy x xy x thay vo phng trnh 2 2 2 1x y y ta c

22 2 2 2 1 2 2 2 3 0x y xy y x x y x y


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1 0

1 3 03 0

x yx y x y

x y

Vi 1 0 1x y x y ta c: 1 1 0 1 0x x x x y

Vi 3 0 3x y x y ta c

23 1 0 2 1 0 1 2x x x x x x y

Vy hphng trnh c nghim ;x y l 1;2 , 1;0

Bi 3.

a) 1

x

R x x

b) Ch :

21 3

1 0 0; 12 4

x x x x x

2

1 2 1 11 1

1 1 1

x x x x xxR

x x x x x x

Bi 4. Gi sngi ca tl x (ngi) (iu kin : , 2x x

)

Stin dnh mi ngi sng:72000

x(ng)

S tin mi ngi phi ng khi tgim i hai ngi :72000

2x

Ta cphng trnh72000 72000

30002x x

72000 72000 2 3000 2x x x x

2 82 48 06

xx x

x


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Theo iu kin ta nhn 8x . Vy tc 8 ngi.

Bi 5.

a)2

2 2

AC aKC a .

2 2 2 3

33 3

aAB AK a .

b) V 0120AHC AOC tgic AHOC ni tip.

0

0180 302

AOCOHC OAC

c) Gi E l im chnh gia cung AC suy ra E l tm g trn ngoi tip tamgic AOC.

V 0

120AIC AOC I thuc ng trn ngoai tip tam gic AOC.Bn knh ng trn ngoai tip tam gic HIO l bn knh ng trn ngoi tiptam gic AOC l EO.

Ta c :6

2 33

aAC a EO EO


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ton khng chuyn 2012- 2013.

Bi 1(2 im). Cho phng trnh

3 4 1 0 1x x x m

a) Gii phng trnh (1) khi 33m .b) Tm m phng trnh (1) c ng hai nghim phn bit 1 2,x x tha

6 61 2 82x x .

Bi 2(2 im).

a) Gii phng trnh 2 7 3 5 1x x .

b) Gii phngtrnh2

2

2 1 2 51 1

510 2

x xy

xy y

Bi 3(2 im).

a) Rt gn biu thc2 2

2 2 2 2

a b abT

ab a b ab a b

vi

, 0a b

, 1a

.

Tm gi trln ca T khi a l stnhin v 1a .

b) Tm 3 stnhin lin tip bit tng 3 tch ca tng cp skhc nhau cachng l 1727.

Bi 4(1 im).

Tng kt hc k2, trng trung hc c sN c 60 hc sinh khng t hc sinhgii trong c 6 em tng t hc sinh gii hc k1; shc sinh gii ca hc k2

bng40

37shc sinh gii ca hc k1 v c 8% shc sinh ca trng khng t

hc sinh gii hc k1 nhng t hc sinh gii hc k2. Tm shc sinh gii hck2 ca trng bit rng shc sinh ca trng khng thay i trong sut nmhc.


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Bi 5(3 im). Cho hnh thang ABCD (AB // CD) ni tip ng trn (C) tm O,

bn knh R v c 0 0105 , 30DAB ACD .

a)

Tnh

DB

DCv AB theo .R

b) Tip tuyn ca (C) ti B ct cc ng thng DO, DA ln lt ti M, N.

TnhMN

MD

c) Gi E l trung im ca AB, tia DE ct MN ti F. TnhBF

BC

Hng dn gii.

Bi 1.

a) Thay 33m phng trnh trthnh 3 4 32 0x x x

t 0t x x phng trnh trthnh

2

84 32 0

4

t nt t

t l

Vi 8 8 4t x x x

Vy phng trnh c 1 nghim 4.x

b) t 0t x x , phng trnh (1) trthnh 2 4 1 0 2t t m

Phngtrnh 1 c ng hai nghim phn bit 2 c hai nghim khng m

phn bit

' 0 3 0

0 4 0 1 3

0 1 0

m

S m

P m

Ta c : 24 4 2 26 6 4 4

1 2 1 1 2 2 1 2 1 2 1 2 1 22 2x x x x x x t t t t t t t t


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Ta c phng trnh 2 2

16 2 1 2 1 82m m

2 230 56 0

28

mm m

m

So vi iu kin nhn 2m .

Vy gi trm cn tm l 2m

Bi 2.

a) Gii phng trnh 2 7 3 5 1x x

iu kin : 2 7 0 7 53 5 0 2 3x x

x

Vi iu kin phng trnh tng ng vi

2 7 1 3 5x x

5 11 2 3 5x x

2

25 122 141 0x x

3

47

25

x

x

So vi iu kin, nhn47

25x

Vy phng trnh c 1 nghim47

25x

b) Gii hphng trnh


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2 2

2 2

2 1 2 5 2 1 2 5

1 1 15 2 2 5 1

10 2 5

x xy x xy

xy y xy y

22

2 2

2 1 2 52 1 2 5

510

5 5

x xyx xy

y xx y

y x

Vi 5y x ta c: 2 2 21 5

2 5 1 2 5 11 5

x yx x x

x y

Vi 5y x ta c: 2 2 21 2 5

2 5 1 2 5 01 2 5

x x x

(v l)

Vy hphng trnh c nghim ;x y l 1; 5 , 1; 5

Bi 3.

a)1

1

aT

a

1 2 21 1 3

1 1 1

aT

a a

V1 1

, 1 2 1 1 11 1

a a a aa

3 2T a

Vy T t gi trln nht l 3 khi 2a

b) Gi 3 stnhin lin tip l 1, , 1a a a .

Theo githit ta c : 1 1 1 1 177 24a a a a a a a

Bi 4. Gi shc sinh gii hc k 2 ca trng l x (hc sinh) ( *x )


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Theo githit lp c phng trnh 37 8

6 60 24040 100

x x x x

Bi 5.

a) V ABCD l hnh thang ni tip ng trn nn ABCD l hnh thang cn

0 0105 ; 30ABC DAB BDC ACD .

Xt tam gic DBC c

0 0 0 0 0105 105 105 30 75DBC ABD BDC

0 0 0 0

180 180 105 75BCD ABC

Do 075DBC BCD nn tam gic DBC cn ti D nn 1DB

DC

Ta c: 0 0 0 0 0360 360 2.30 2.105 90AB AD DCB


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b) Xt tam gic AOD c : OA OD R v 02 60DOA ACD

AOD u 060ADO

Ta li c: ; / /OA OB MN OB OA MN

060DMN DOA (ng v)

Do 060DMN ADO Tam gic DMN u nn 1MN

MD

c)

Ta c tam gic OAB vung cn ti O, E l trung im AB nn OE EA v DA DO (tam gic ADO u) nn suy ra DE l trung trc ca OA

DE OA m / /OA MN DE MN .

Tam gic EFB c EF FB v 01

452

EBF sd AB nn tam gic EFB vung

cn ti F nn2

EBFB

Tam gic OEB vung cn ti E nn 2OB EB BC (tam gic OBC u)

Vy12

22

EB

FB

BC EB

090 2AOB AB R


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Ton khng chuyn 2013 2014

Bi 1(2 im).

a) Gii phng trnh 1 2x x .b) Tm chiu di ca mt hnh chnht c chu vi l a (mt), din tch l a (mt

vung) v ng cho l 3 5 (mt).

Bi 2(2 im). Cho phng trnh 21 5 1 0 1x x x m

a)

Gii phng trnh (1) khi 1m .b) Tm m phng trnh (1) c ba nghim phn bit 1 2 3, ,x x x tha

2 2 21 2 3 1 2 3 1 2 2 3 3 1 31x x x x x x x x x x x x .

Bi 3(2 im)

a) Vi 0 b a , hy rt gn biu thc

1 2 1 2: 111 1

a b a b a bP b a ba a b a a b

b) Gii hphng trnh

2 1 1

2

x yx y

x y xy

Bi 4(1 im). C hai vi nc A, B cng cung cp cho mt hcn nc v vi C(t st y h) ly nc thcung cp cho hthng ti cy. ng6 gi, hai viA, B c m, n 7 givi C c m; n 9 gith ng vi C v vi B; n 10gi45 pht th hy nc. Ngi ta thy rng nu ng vi B ngay tu th

phi ng 13 gihmi y. Bit lu lng vi B l trung bnh cng ca vi A vvi C, hi mt mnh vi C tho cn hnc y trong bao lu ?

Bi 5(3 im).


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Cho tgic ABCD ni tip trong ng trn ng knh AC, AC = 2a v tam gicABD u. Gi M, N ln lt l trung im ca AB v AD.

a) Tnh BC, CN theo a.

b)

Gi H l trc tm tam gic CMN, MH ct CN ti E, MN ct AC ti K.Chng minh 5 im B, M, K, E, C cng thuc mt ng trn (T). ngtrn (T) ct BD ti F ( F B ), tnh DF theo a.

c) KF ct ME ti I. Chng minh KM tip xc vi ng trn ngoi tip tamgic MIF Tnh gc IND.

Hng dn gii.

Bi 1.

a)

2 2

2 0 21 2

5 3 01 2

x xx x

x xx x

25 13

5 13 5 132

2 2

x

xx x

Vy phng trnh c mt nghim5 13

2x

b) Gi di hai cnh ca hnh chnht l ,x m y m , iu kin , 0x y

Chu vi hnh chnht : 2 x y a

Din tch hnh chnht .x y a

ng cho hnh chnht: 2 2 2 23 5 45x y x y

Ta c: 2 22 2 45 2 45 0 4 4 180 0x y x y xy x y xy

22 2 188 180 0 4 14

10

aa a a

a


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Ta c:9

18

x y

xy

, ,x y l nghim ca phng trnh 2

69 18 0

3

XX X

X

Vy chiu di ca hnh chnht l 6 m.

Bi 2.

a) Khi 1m , ta c phng trnh 21 5 2 0x x x

2

00 1

11 0 5 33

5 33 5 3325 2 0

2 2

xx x

xx

xx xx x

Vy phng trnh c hai nghim5 33

1;2

x x

b)

2

2

0

11 5 1 0 1

5 1 0 2

x

xx x x m

x x m

Gisphng trnh (1) c ba nghim phn bit 1 2 31, ,x x x trong 2 3,x x l

nghim ca phng trnh (2).

Theo nh l viet ta c : 2 3 1 25, 1x x x x m

Ta c: 2 2 21 2 3 1 2 3 1 2 2 3 3 1 31x x x x x x x x x x x x

2 22 3 2 3 2 3 2 3 29x x x x x x x x

2

2 3 2 3 19 25 1 19 7x x x x m m


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Thay 7m vo phng trnh (1) ta c

22

0 1

11 5 6 0 2

35 6 0

x x

xx x x x

xx x

Tha phng trnh (1) c 3 nghim phn bit.

Vy 7m l gi trcn tm.

Bi 3.

a)1

a bP

b

b)

2 1 1

2

x yx y

x y xy

iu kin: 0xy

Tphng trnh u ca h

2 21 1 1

0x y

x y x y x y x yx y xy xy

1

x y

x yxy

Vi x y

thay vo phng trnh thhai ta c2 2

2 2

x

x x

Hphng trnh c nghim ;x y l 2; 2 , 2; 2


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Vi1

x yxy

thay vo phng trnh thhai ta c:

212 2 1 0 1 1xy xy xy xy x y

xy

1y x

Ta c: 21 5 5 1

2 21 1 2 1 0

1 5 1 5

2 2

x y

x x x x

x y

Phng trnh c nghim ;x y l5 1 5 1 5 1 5 1

; , ;2 2 2 2

Vy phng trnh cho c 4 nghim ;x y l

5 1 5 1 5 1 5 12; 2 , 2; 2 , ; , ;2 2 2 2

Bi 4. Gi x (gi) l thi gian vi A chy y b

y (gi) l thi gian vi B chy y b

z (gi) l thi gian vi C tho cn b

iu kin. , , 0x y z

Nh vy trong 1 gith vi A chy c1

x(h), vi B chy c

1

y(h) v vi C

tho c1

z(h).

Lp c hphng trnh


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19 3 21

4

7 21

1 1 1 1

2

x y z

x z

y x z

6, 8, 12x y z

Bi 5.

a)

BC a ,

2 2

22

3 33

2 2 34 2 2 4 2 4

aa

MN a a aCN

2 23 25 28 7

16 16 4 2

a aa a


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b) B, M, K, E, C cng thuc ng trn ng knh MC.

V KFDN l hnh bnh hnh nn1 3

4 4

aDF KN BD

c)

CMN cn ti C, ta c :

KME KCE KCM KFM

T chng minh c KM l tip tuyn ca ng trn ngoi tip tam gic MIF

v 090IND


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Ton khng chuyn 2014 2015.

Bi 1(2 im).

a) Gii phng trnh 23 3 9 4 5 3x x x x .

b) Tnhx

ybit rng 1, 0x y v

23 3

2 2 3 4

1 4 16

1 4 1

x y x y x

x x y xy y

Bi 2(2 im).

a) Gii hphng trnh 2 2

2

2 9 7 15 0

9 7 8

x y x y

x y

b) Cho hnh thoi ABCD c din tch l 18 3 (mt vung), tam gic ABD u.Tnh chu vi hnh thoi v bn kinh ng trn ngoi tip tam gic ABC.

Bi 3(2 im). cho phng trnh 2

3 2 1 0 13

mx m x mx

a) Gii phng trnh (1) khi 1m .b) Tm m phng trnh (1) c hai nghim phn bit 1 2,x x sao cho

21 2 221 7 2 58x m x x .

Bi 4(1 im).

a) Gi ,2

a bx y ab

ln lt l trung bnh cng v trung bnh nhn ca

hai sdng a, b. Bit rng trung bnh cng ca ,x y bng 100. Tnh S a b .


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b) Gishai i lng ,x y tlnghch ( ,x y lun dng). Nu x tng %a th

y gim %m . Tnh m theo .a

Bi 5(3 im). Hnh vung ABCD c 2AB a . AC ct BD ti I. Gi (T) l ng

trn ngoi tip tam gic CID, BE tip xc vi (T) ti E (E khc C), DE ct AB ti F

a) Chng minh tam gic ABE cn. Tnh AF theo a.b) BE ct AD ti P. Chng minh ng trn ngoi tip tam gic ABP tip xc

CD. TnhAP

PD.

c) AE ct (T) ti M (M khc E). Tnh AM theo a.

Hng dn gii.

Bi 1.

a)Gii phng trnh 23 3 9 4 5 3x x x x .

iu kin: 3 0 3x x

Phng trnh cho tng ng vi 2 23 3 9 16.5 3x x x x

23 3 3 9 80 0x x x x

43 1 0x x

4

33 0

11 0

1

xx

xx

x

(tha iu kin)

Vy phng trnh c 3 nghim 1; 1; 3x x x

b)

Ch : 1, 0 0x

x yy

v 4 1 1 4.1 1 1 3 1 0x


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Ta c :

23 3

2 2 3 4

1 4 16

1 4 1

x y x y x

x x y xy y

2 2

2 2 2

1 4 16

1 4 1

x y x y x xy y x

x y x xy y

22 2

26 7 7

x y x x

y yy

Bi 2.

a)

22 2

22

2

7 0

2 02 9 7 15 0

9 7 15 09 7 0

9 7 8

y

x yx y x y

x yx y

x y

22

22

7

9 7 152 01 2

9 7 8 9 7 8

y

x yx y

x y x y

Vi iu kin 7y

Gii (1)

Ta c: 22

2 2

9 7 02 0

9 7 8 9 7 8

x yx y

x y x y


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t 2 9 3, 7 0u x v y ta c h

2 2 0 8 400 48 8

u v u v u v uu v

u v vu v u v

(nhn)

Do ta c :

2 79 4

7

xx

x

v 7 4 9y y

Hnghim ;x y l 7;9 , 7;9

Gii (2)

2

2

9 7 15

9 7 8

x y

x y

t 2 9, 7 3, 0u x v y u v

Ta c h15 5

8 3

uv u

u v v

(nhn) hoc

3

5

u

v

(nhn)

Vi 5, 3u v ta c: 24

9 54

xx

x

v 7 3 2y y

Hc nghim ;x y l 4;2 , 4;2

Vi 3; 5u v ta c: 2 9 3 0x x v 7 5 18y y

Hc nghim ;x y l 0;18

Vy h cho c nghim ;x y l 7;9 , 7;9 , 0;18 , 4;2 , 4;2 .

b)


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Tam gic ABC u nn3

3

2

ABAO AC AB v BD AB

1 1. 3. 18 3 6

2 2ABCDS AC BD AB AB AB

Vy chu vi hnh thoi l 24 v bn knh ng trn ngoi tip tam gic ABC l 6.

Bi 3.

a)Khi 1m , phng trnh (1) trthnh2

4 30

3

x x

x

1 30 1

3

x xx

x

(iu kin 3x )

phng trnh c mt nghim l 1x

b)gisphng trnh (1) c hai nghim phn bit 1 2,x x tha mn hthc

21 2 221 7 2 58x m x x

V 2x l nghim ca 2 22 2 2 2 21 3 2 1 0 2 3 1mx m x m m x x x

Do ta c: 21 2 2 1 2 1 217

21 7 2 58 21 7 3 1 587

x m x x x x x x


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V 1 2,x x l nghim ca (1) 1 2,x x l nghim ca phng trnh

2 3 2 1 0mx m x m

Ta c: 1 2 3 17 77 8

mx x mm

Thay7

8m vo (1) ta c phng trnh

22

7 17 6 27 17 68 8 8 0 0 3

3 37

xx xx x

x x x

Vi7

8m th (1) c hai nghim phn bit.

Vy7

8m l gi trcn tm.

Bi 4.

a)

2

100 200 400 2002 2

x y a bab a b a b

(v 0a b )

b)theo githit x tng %a th y gim %m . V ,x y l hai i lng tlnghch

nn ta c100

. . 1 1 1100 100 100 100 100

a m a m ax x y y xy m

a

Bi 5.


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a) V BE, BC l tip tuyn ca (T) nn BE = BC = BA ABE cn ti B.

Ta c tgic CEFD ni tip, Do ta suy ra c

AFD FDC BEC BFC , do chng minh c

2

ABAFD BFC AF BF a

b) Gi K l tm ng trn ngoi tip ABP K l trung im ca BP.O l tm ng trn ngoi tip CDI O l trung im ca CD.

trong hnh thang vung PDCB c KO l ng trung bnh nn KO CD

Mc khc2 2 2

PD BC PE EB PDKO

CD tip xc vi ng trn ngoi

tip tam gic ABP.

Ta c:2 2

2.

2 2

OE a aOE PE BE PE

BC a


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Do 3

2 32 2 2

a a a APPD AP a

DP

c) Ta c : 045IEF ICD IAF tgic AEIF ni tip

090AE EI IEM IM l ng knh ca (T)

T tnh c 10AM a


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Ton khng chuyn 2015- 2016.

Bi 1(2 im).

a) Gii phng trnh 2 29 2 9x x x x .

b) Gii hphng trnh

22 2 2 2

2 2

4 4 4 5

3 2 5

x y x y

x y

Bi 2(1,5 im) Chophng trnh

2 3

0 11

x m x m

x

a) Tm m phng trnh (1) c hai nghim phn bit 1 2,x x .

b) Tm m 2 2 21 2 1 25 14 30 4x x x x m m

Bi 3(1,5 im)

a) Rt gn biu thc

3 3 36 5

: 0; 9, 2593 3 3

x x xQ x x x

xx x x x

.

b)

Tm x 0Q

Bi 4(2 im)

a) Cho tam gic vung. Nu ta tng di cc cnh gc vung thm 3 cm th

din tch tng 233cm , nu gim di mt cnh gc vung 2 cm v tng

di cnh gc vung kia 1 cm th din tch gim 22cm . Hy tnh di cc

cnh ca tam gic vung.

b)

Bn An dnh trong khong thi gian tngy 1/3 n ngy 30/4 mi ngysgii 3 bi ton. Thc hin ng khoch mt thi gian, vo khong cuithng 3(thng 3 c 31 ngy) th A bbnh, phi nghgii ton nhiu ngylin tip. Khi phc hi, trong tun u An gii 16 bi ton; sau An cgng gii 4 bi mt ngy v n 30/4 th An cng hon thnh khoch nh. Hi bn An nghgii ton t nht bao nhiu ngy ?


35/38



Bi 5(3 im). Hnh bnh hnh ABCD c 060ADC v tam gic ACD nhn.ng trn tm O ngoi tip tam gic ACD ct cnh AB ti E (E A ), AC ct DEti I.

a)

Chng minh tam gic BCE u v OI CD .b) Gi K l trung im BD, KO ct DC ti M. Chng minh A, D, M, I cng

thuc mt ng trn.

c) Gi J l tm ng trn ngoi tip tam gic ABC. TnhOJ

DE

Hng dn gii.

Bi 1.

a) Gii phng trnh 2 29 2 9x x x x

iu kin: 2 0 2x x .

Phng trnh cho tng ng vi 2 9 2 0x x x

2 9 0

2

x

x x

Vi 2 9 0 3x x (loi) hoc 3x (nhn)

Vi2 2

00 0

2 112 2 0

2

xx x

x x xxx x x x

x

(nhn).

Vy phng trnh c hai nghim 3; 1x x

b)

2

2 2 2 2

2 2

4 4 4 5

3 2 5

x y x y

x y


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phng trnh u 2 2 2 24 5 4 1 0x y x y

v 2 24 1 0x y nn 2 25 5x y

Ta c hphng trnh2 2 2

2 2 2

5 5 1

3 2 5 1

x y x

x y y

Vy hphng trnh c cc nghim ;x y l 1;1 , 1;1 , 1; 1 , 1; 1

Bi 2.

a)phng trnh c hai nghim phn bit

2 3 1

12 12

1 3 2

m m m

m m

m m

b) 22 3

031

x mx m x m

x mx

Vi iu kin cu a), phng trnh c hai nghim 2 , 3x m x m

Thay vo 2 2 21 2 1 25 14 30 4x x x x m m ta c:

2 2 22 3 5 2 3 14 30 4m m m m m m

2 16 5 0

5

mm m

m

So vi iu kin loi m =1 v nhn m =5.

Vy 5m l gi trcn tm.

Bi 3.

a)12

5

xQ

x


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b)Ta c 0 5 0 25Q x x

so vi iu kin c p sl 0 25, 9x x

Bi 4.

a)Gi ,x y l di cc cnh g vung 0, 0x y . Ta c

3 3 66 12

72 1 4

xy x y x

yxy x y

b)gi x l sngy An gii ton trc khi nghbnh, 0 31x ; y l sngy An

nghlm ton. ,x y .

Ta c: 3 4 61 7 16 61.3 4 49x x y x y

Do 31x nn 4 18y , m y nn 5y .

Khi 5y th 29x (tha mn). Vy An nght nht 5 ngy.

Bi 5.

a) Ta c : 060BEC ADC v 060EBC ADC suy ra tam gic BCE u.


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ABCD l hnh bnh hnh c 0 060 120ADC DAB

Ta c : 0 0180 60ECD DAB

Do hnh thang AECD l hnh thang cn, suy ra IC = ID

M OC = OD nn IO l trung trc ca CD nn IO CD

b)K l trung im BD nn K l trung im AC

OK l trung trc ca AC.

M thuc OK nn MA = MC MAC cn ti M.

Do CAM MCA IDM tgic ADMI ni tip.

c) Ta c OJ AC ti K. Hn na do ABC ADC KO KJ

Ta li c DE AC

Nn 0 1

cot cot 603

OJ KJ KJO

DE KC

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