On Thi Lop 10 Toan VuVanBac

8/12/2019 On Thi Lop 10 Toan VuVanBac

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S GIO DC V O TO NAM NH

TRNG THPT A NGHA HNG NAM NH

TI LIU N THI VO LP 10 MN TON

Thc hin V Vn Bc Website: http:/ / parksungbuyl.wordpress.com/

---------- NGH A HNG NGY 8 THNG 5 NM 2013 ----------

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Ti li u n thi vo l p 10 mn Ton by V Vn Bc

VN 1. RT G N BIU THC C CHA CN

A. PH NG PHP GII TON

Bi ton 1.1 Cho bi u thc2

1 1 x x x

P x x x

vi 0, 1. x x

a) Rt g n biu thc P b) Tm x khi 0. P

( Trch thi tuyn sinh v o l p 10 tnh Nam nh nm 2011)

Li gii.a) Vi 0, 1 x ta c

3

11 1 1

1 1 1 1

x x x x x x x x x

P x x x x x x

1 1

11

x x x x x x x x

x x

2 . x x x x x

Vy vi 0, 1 x th 2 . P x x b) Vi 0, 1 x ta c

0 2 0 2 0 P x x x x

00 0

422 0

x x x

x x x

i chiu vi iu kin 0, 1 x x ta thy hai gi tr ny u tha mn.Vy vi 0 P th 0, 4. x x

NHNG IM CN LU KHI GI I TON

K nng cng nh cch gii chung cho d ng ton nh cu a t iu kin thch hp,nu bi nu iu kin xc nh th ta v n phi ch

ra trong bi lm c a mnh nh li gii nu trn. a phn cc bi ton dng ny, chng ta thng quy ng mu, xong ri tnh ton

rt gn t thc v sau xem t thc v mu thc c tha s chung no haykhng rt gn tip. Trong bi ton trn th khng quyng mu m n gin biu thc lun. Khi lm ra k t qu cui cng, ta k t lun ging nh trn.

i vi dng ton nh cu b Cch lm trn l in hnh, khng b tr im. Ngoi cu hi tm x nh trn th ngi ta c th hi: cho x l mt hng s no

bt rt gn P , gii bt phng tr nh, tm gi tr ln nht nh nht, tm x P c

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gi tr nguyn, chng minh mt bt ng thc. Nhng thng th ngi ta s hinh sau: tm x P c gi tr no (nh v d nu trn), cho x nhn mt gi trc th tnh P .

MT S CU HI M CHO B I TON

Cu h i m 1.Rt gn P khi 3 2 2. x

Ta c 2 2 23 2 2 1 2.1. 2 ( 2) (1 2) x

Khi , vi 0, 1 x x th 2(1 2) 1 2 x Do 2 3 2 2 2(1 2) 3 2 2 2 2 2 1. P x x Vy vi 3 2 2 x th 1. P

Cu h i m 2.Tm gi tr nh nht ca P V i 0, 1 x ta c 2 22 ( ) 2 1 1 ( 1) 1 P x x x x x

V 1 x nn 2( 1) 0 x 2( 1) 1 1 x Vy vi 0, 1 x th P khng c gi tr nh nht.

Trong loi cu hi ny, ta cn ch n iu kin xc nh. Chng hn vi iu kin4 x ta rt gn c P x x th ta s khng lm nh trn m s lm nh sau

Vi 4 x ta c 2 ( 2) P x x x x x x V 4 2 0, 2 0 ( 2) 0 2 2 x x x x x x x Vy min 2 P , du bng xy ra khi v ch khi 4 x (tha mn iu kin).

Cu h i m 3.Chng minh rng 1 P th ta lm nh trn nhngk t lun l

1. P Cu h i m 4.Tm s nguyn x P c gi tr nguyn.

V d tr n, ta c 2 P x x , th thng bi s khng hi n nghim nguyn. Chng

hn vi iu kin 1 x ta rt gn c 31

x P , bi hi: tm s nguyn x P nhn

gi tr nguyn th ta lm nh sau

Vi 1, ta c 3 3( 1) 3 331 1 1

x x P

x x x

T vi x l s nguyn, 3 33 3 ( 1)1 1

P x

x x

M

Tng ng vi 1 x l c ca 3, m c ca 3 l 3; 1;1;3 ( 1) 3; 1;1;3 x M 1 1 2 1 3 2 x x x x (tha mn iu in) K t lun: vy 2 x l gi tr cn tm.Ta xt thm mt bi ton na l mt cu trong chung chuyn L HngPhong Namnh nm 2011.

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Bi ton 1.2 Cho bi u thc 3 1 1 1:1 1

x P

x x x x

vi 0, 1. x x

a) Rt g n biu thc B b) Tm x 2 3. P x

( chung Chuy n L H ng Phong Nam nh nm 2011) Li gii.a) Vi 0, 1 x ta c

3 1 1( 1)( 1) ( 1)( 1) x x

B x x x x x x

3 1 1( 1).( 1)( 1)

x x x x

x x

(2 2) 2 ( 1) 2 .

1 1

x x x x

x x

Vy vi 0, 1 x th 2 . P x b) Vi 0, 1 x v 2 P x ta c

2 3 4 34 3 0

3 3 0( 1) 3( 1) 0

( 1)( 3) 0

1 0 1 193 0 3

P x x x

x x

x x x

x x x

x x

x x x

x x x

K t hp vi iu kin nu trn th ch c 9 x tha mn bi ton.

B. CC BI TON RN LUY N

Bi 1 : Cho biu thc6

532

aaa

a P

a21

a) Rt gn P . b) Tm gi tr caa 1. P

Bi 2 : Cho biu thc P =

652

32

23:

11

x x

x

x

x

x

x

x

x

a) Rt gn P . b) Tm gi tr ca x 0. P

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Bi 3: Cho biu thc P =

13231:

198

131

131

x

x x

x

x x

x

a) Rt gn P .

b) Tm cc gi tr ca x 6.5

P


12

11:

11

aaaa

a

aaa

a) Rt gn P. b) Tm gi tr caa 1. P c) Tm gi tr ca P nu 3819 a


a

a

aa

a

a

a

aa

1

1.1

1:1

)1( 332

a) Rt gn P b) Xt du ca biu thc ( 0,5). M a P

Bi 6 : Cho biu th P =

122

1211:1

122

121

x x x

x x

x x x

x x

a) Rt gn P

b) Tnh gi tr ca P khi 3 2 2.2

x


11:

11

12

x x

x x x x x

x

a) Rt gn P b) Tm x P 0


a

aa

aaa

a

a11.

112 3

3

a) Rt gn P

b)

Xt du ca biu thc P a1

Bi 9 : Cho biu thc 1 1 2 1 2:11 1

x x x x x P

x x x x x

a) Rt gn P b) Tnh gi tr ca P vi 7 4 3 x c) Tnh gi tr ln nht ca a . P a

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Bi 10: Cho biu thcP =

a

a

aaa

a

aa11.

11

a) Rt gn P. b) Tma 7 4 3. P

Bi 11 : Cho biu thcP =

1

322:

933

332

x

x x x

x

x

x

x

a) Rt gn P

b) Tm x 12

P

c) Tm gi tr nh nht ca P


32

23

69:1

93

x

x

x

x

x x

x x

x x

a) Rt gn P b) Tm gi tr ca x P < 1

Bi 13 : Cho biu thcP =332

123

321115

x

x

x

x

x x

x

a) Rt gn P

b) Tm cc gi tr ca x 12

P

c) Chng minh 2.3

P

Bi 14 : Cho biu thcP = 22

442

m xm

m x

x

m x

x vi m > 0

a) Rt gn P b) Tnh x theo m P = 0.c) Xc nh cc gi tr ca m x tmc cu b tho mn iu kin 1. x

Bi 15: Cho biu thcP = 121

2

a

aa

aa

aa

a) Rt gn P b) Bit 1a hy so snh P vi P c) Tm a P = 2d) Tm gi tr nh nht ca P


1

111:1

111

ab

aab

ab

a

ab

aab

ab

a

a) Rt gn biu thc P.

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b) Tnh gi tr ca P nua = 32 v b =3113

c) Tm gi tr nh nht ca P nu 4ba

Bi 17: Cho biu thcP =

11

11111

aa

aa

aa

aaaa

aaaa

a) Rt gn P b) Vi gi tr no ca a th P = 7c) Vi gi tr no caa th 6. P


11

11

21

2

2

a

a

a

a

a

a

a) Rt gn P b) Tm cc gi tr ca a P 0 c) Tm cc gi tr ca a P 2

Bi 19 : Cho biu thcP = ab

abba

ba

abba .42

a) Tmiu kin P c ngha. b) Rt gn P c) Tnh gi tr ca P khi a = 32 v b = 3

Bi 20 : Cho biu thcP = 2 1:1 1112

x

x x x x

x x x

a) Rt gn P b) Chng minh rng P > 0 vi x 1


121:

11

12

x x

x

x x x

x x

a) Rt gn P b) Tnh P khi x = 325

Bi 22 : Cho biu thcP = x x x

x

x 241:

242

423

21:1

a) Rt gn P b) Tm gi tr ca x P = 20

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Bi 23 : Cho biu thc P = y x

xy y x x y

y x

y x

y x

233:

a) Rt gn P b) Chng minh P 0

Bi 24: Cho P =

baba

ba

bbaa

ab

babbaa

ab

ba:31.31

a) Rt gn P b) Tnh P khi a = 16 v b = 4

Bi 25 : Cho biu thc P =12

.1

21

121

a

aa

aa

aaaaaaa

a) Rt gn P

b) Cho P =61

6 tm gi tr ca a

c) Chng minh rng 2.3

P

Bi 26: Cho biu thc:P=

35

53

15225:1

255

x

x

x

x

x x

x x

x x

a) Rt gn P b) Vi gi tr no ca x th 1. P

Bi 27 : Cho biu thc P = bababaa

babbaaa

babaa

222.1:133

a) Rt gn P b) Tm nhng gi tr nguyn ca a P c gi tr nguyn


12

21:1

11

a

a

a

a

aa

a) Rt gn P

b) Tm gi tr ca a 1.6 P

Bi 29 : Cho biu thcP =33

33

:112.11 xy y x

y y x x y x

y x y x y x

a) Rt gn P b) Cho x.y =16. Xc nh x, y P c gi trnh nht.

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Bi 30 : Cho biu thc P = x

x

y xy x x

x

y xy

x

11.

222

2

3

a) Rt gn P b) Tm tt c cc s nguyn dng x y = 625 vP 0,2.

VN 2 . PHNG TR NH B C HAI MT N

A. PHNG PHP GII TON

Xt phng tr nh 2 0ax bx c via khc 0, bit thc 2 4 .b ac H thc Viet i vi phng tr nh bc hai

1 2 1 2;b c

x x x xa a

Nu 0ac th PT c 2 nghim phn bit. PT c nghim 0. PT c nghim kp 0. PT c 2 nghim phn bit 0.

PT c 2 nghim phn bit tri du1 2

00 x x

PT c 2 nghim dng phn biT 1 21 2

00

0

x x

x x

PT c 2 nghim m phn bit 1 21 2

00

0 x x

x x

T nhng tnh cht quan trng nu trn, ta s gii c mt dng ton v PT tr ng phng.

Xt phng tr nh 4 2 0ax bx c (i) via khc 0. t 2 0t x , ta c2 0.at bt c (ii)

PT (i) c 4 nghim phn bit khi v ch khi (ii) c 2 nghim dng phn bit. PT (i) c 3 nghim phn bit khi v ch khi (ii) c 1 nghim dng v 1 nghim

bng 0. PT (i) c 2 nghim phn bit khi v ch khi (ii) c duy nht mt nghim dng. PT (i) c 1 nghim k hi v ch khi (ii) c duy nht mt nghim l 0.

Sau y chng ta s xt mt s bi ton thng gp mang tnh cht in hnh.

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Bi ton 2.1 Cho phng tr nh 2( 1) 4 4 1 0.m x mx m (1)a) Hy gi i phng tr nh trn khi 2m b) Tm m phng tr nh c nghi m. c) Tm m phng tr nh c hai nghi m phn bit. Khi hy tm mt biu thc

lin h c lp gia cc nghim ca phng tr nh.d) Tm m phng tr nh c hai nghi m phn bit 1 2, x tha mn

1 2 1 2 17. x x x x e) Tm m phng tr nh c hai nghi m dng phn bit.

f) Tm m phng tr nh c hai nghi m m phn bit. g) Tm m phng tr nh c hai nghi m phn bit tri du. h) Tm m khi 1 2 2 7 x x , vi 1 2, x l hai nghi m ca phng tr nh.i) Tm m phng tr nh c 2 nghi m phn bit tha mn nghi m ny bng 2 ln

nghim kia. Li gii.a) Khi 2m thay vo (1) ta c2 8 9 0 x x (2)

PT ny c ' 16 9 7 0 Khi (2) c hai nghim 1 24 7; 4 7 x x Vy vi 2m th PT cho c tp nghim l 4 7;4 7 .S b) lm cu hi ny, ta s chia thnh hai trng hp

TH1: Khi 51 5 4 0 14

m x x m tha mn.

TH2: Khi m khc 1, PT (1) l PT bc hai. Xt2 2 2' 4 ( 1)(4 1) 4 (4 3 1) 3 1m m m m m m m

PT (1) c nghim khi 1' 0 3 1 03

m m

Tm li, vy vi 13

m th PT cho c nghim.

c) PT (1) c 2 nghim phn bit khi11 1

1' 0 3 1 03

mm m

m m

Khi , p dng h thc Viet ta c

1 24 4( 1) 4 44

1 1 1

m m x x

m m m

1 24 1 4( 1) 5 54

1 1 1m m

x xm m m

Do 1 2 1 24 55 5 4 4 5 4 1

1 1 x x x x

m m

Vy biu thc cn tm l 1 2 1 25 4 1 . x x x

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d) PT (1) c 2 nghim phn bit khi11 1

1' 0 3 1 03

mm m

m m

p dng h thc Viet ta c1 2 1 24 4 1

;1 1m m

x x x xm m

Khi vi 11,3

m m ta c

1 2 1 24 4 1 4 4 117 17 17

1 1 1m m m m

x x x xm m m

8 1 17 8 1 17 17 9 18 21

mm m m m

m (tha mn K)

Vy 2m l gi tr cn tm.

e) PT (1) c 2 nghim dng phn bit khi v ch khi 1 21 2

' 0

00

x x x x

1' 03

m

1 2

14 10 0 (4 1)( 1) 0 114

mm

x x m mm m

1 2

140 0 4 ( 1) 001

mm x x m m

mm

Vy PT cho c 2 nghim dng phn bit khi 1 11 or .3 4

m m

f) PT (1) c 2 nghim m phn bit khi v ch khi 1 21 2

' 00

0 x x

x x

n y ta lm tng t nh cu e.

g) PT (1) c 2 nghim phn bit tri du khi v ch khi1 2

' 00 x x

n y ta lm tng t nh cu e. h) Bnh phng hai v v lm tng t nh cu d, ch

2 2 21 2 1 2 1 2 1 24 . x x x x x x x x

i) K PT (1) c 2 nghim phn bit: 11, .3

m m

T gi thit bi ton, ta c: 1 2 2 1 1 2 2 12 or 2 2 2 0 x x x x x x x x

22 21 2 1 2 1 2 1 25 2 0 9 2 0 x x x x x x x x

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p dng h thc Viet ta c1 2 1 24 4 1;

1 1m m

x x x xm m

, nn

22

29(4 1) 2.16 0 9( 1)(4 1) 32 0

1 ( 1)m m

m m mm m

2 2 236 27 9 32 0 4 27 9 0m m m m m n y cc em lm tip, ch iu kin PT c 2 nghim phn bit.

NHNG IM CN LU KHI GI I TON i vi nhng bi ton c lin quan n h thc Viet, th tac bit quan tm n

K phng tr nh c nghim, tm ra c x, ta phi i chiu K PT cnghim.

Ngoi cc cu hi nh trn ta cn c th hi: tm m thng qua gii bt phngtrnh (tng t nh cu hi d), tm gi tr ln nht nh nht. V d tr n, h s ca

x2

l tham s nn khi p dng Viet ta thy c bin mu, thng ngi ta skhng hi min max bi ny. i vi bi ton m h s ca x2 khng cha tham s th ta c th hi min max

thng qua h thc Viet. Chng hn cho PT2 22( 1) 1 0 x m x m . Tm mPT c 2 nghim 1 2, x x ; khi tm min ca biu thc 1 2 1 22 P x x x x ta cth lm nh sau

dng tmc K PT c 2 nghim1 2, x x l 1m (cc em lm ng k nng nhVD). p dng Viet ta c 21 2 1 22 2; 1 x x m x x m Khi ta c 2 21 2 1 22 1 2(2 2) 4 3 P x x x x m m m m

n y c mt sai lm m a s HS mc phi l phn tch2 24 3 ( 2) 1 1m m m v k t lun ngaymin 1. P

i vi bi ton ny, cch lm trn hon ton sai. Da vo iu kin PT c nghim l1m , ta s tm min ca P sao cho du bng xy ra khi 1.m Ta c

2 24 3 3 3 ( 1) 3( 1) ( 1)( 3) P m m m m m m m m m m Vi 1 1 0, 3 0 ( 1)( 3) 0 0m m m m m P Vy min 0 P , du bng xy ra khi 1m (tha mn K nu).

Bi ton 2.2 Tm m PT 2 4 3 1 0 x mx m (i) c hai nghi m 1 2, x tha mn

1 22 . x x

Li gii.PT (i) c 2' 4 3 1m m , (i) c 2 nghim2 2' 0 4 3 1 0 4 4 1 0

4 ( 1) ( 1) 0 ( 1)(4 1) 011 or .4

m m m m m

m m m m m

m m

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Khi theo h thc Viet ta c1 2 1 24 ; 3 1 x x m x x m (*)

Ta li c 1 21 21 2

22

2 x x

x x x x

+ Vi 1 22 x x k t hp vi (*) ta c

1 2 1 2 1 2

1 2 2 2 22

1 2 2 2 2

2 2 24 2 4 3 4

3 1 2 3 1 2 3 1

x x x x x x x x m x x m x m

x x m x x m x m

T 2 233 44

x m m x , th vo 222 3 1 x m ta c

2 2 22 2 2 2 2 2

92 1 8 9 4 8 9 4 0.4

x x x x x x

n y, cc em lm tip r n luyn k nng. + Vi 1 22 x x ta lm tng t nh trn.

Nh n xt. Bi ton trn, ta th m bi 2 bi l, khi lm nh vy ta khng phi khai phng tc l nu th 2 x bi m th ta s phi khai phng, khng thun li. Ngoi cchlm trn ta cn c th gii nh sau: 1 2 1 2 1 22 2 2 0. x x x x x x T khaitrin ra v dng h thc Viet gii.


Bi 1: Cho phng tr nh 22 2122 m x xm a) Gii phng tr nh khi 12 m b) Tm m phng tr nh c nghim 23 x c) Tm m phng tr nh c nghim dng duy nht.

Bi 2 : Cho phng tr nh 0224 2 mmx xm a) Tm m phng tr nh c nghim 2 x . Tm nghim cn li. b) Tm m phng tr nh 2 c nghim phn bit.c) Tnh 2221 x x theo m.

Bi 3 : Cho phng tr nh 04122 m xm x a) Tm m phng tr nh 2 c nghim tridu b) Chng minh rng phng tr nh lun c 2 nghim phn bit vi mi m c) Chng minh biu thc M = 1221 11 x x x x khng ph thuc vo m.

Bi 4 : Tm m phng tr nha) 0122 m x x c hai nghim dng phn bit b) 0124 2 m x x c hai nghim m phn bit c) 012121 22 m xm xm c hai nghim tri du.

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Bi 5 : Cho phng tr nh 021 22 aa xa x a) Chng minh rng phng tr nh trn c 2 nghim trI du vi mi a b) Gi hai nghim ca phng tr nh l x1 v x2. Tm gi tr ca a 2221 x x t gitr nh nht

Bi 6 : Cho b v c l hai s tho mn h thc 2111 cb

Chng minh t nht mt trong hai phng tr nh sau phi c nghim2

2

00.

x bx c

x cx b

Bi 7: Vi gi tr no ca m th hai phng tr nh sau c t nht mt nghim s chung

2

2

2 (3 2) 12 04 (9 2) 36 0 x m x

x m x

Bi 8: Cho phng tr nh 0222 22 mmx x a) Tm cc gi tr ca m phng tr nh c hai nghim dng phn bit b) Gi s phng tr nh c hai nghim khng m, tm nghim dng ln nht ca

phng tr nh.

Bi 9 : Cho phng tr nh 0142 m x x a) Tmiu kin ca m phng tr nh c nghim b) Tm m sao cho phng tr nh c hai nghim x1 v x2 tho mn iu kin

102221 x x

Bi 10 : Cho phng tr nh 052122 m xm x a) Chng minh rng phng tr nh lun c hai nghim vi mi m b) Tm m ph ng trnh c hai nghim cung du. Khi hai nghim mang du g.

Bi 11 : Cho phng tr nh 0102122 m xm x a) Gii v bin lun vs nghim ca phng tr nh b) Trong trng hp phng tr nh c hai nghim phn bit l 21; x x hy tm mt h

thc lin h gia 21; x x m khng ph thuc vo mc) Tm gi tr ca m 22212110 x x x x t gi tr nh nht.

Bi 12 : Cho phng tr nh 0121 2 mmx xm a) Chng minh phng tr nh lun c hai nghim phn bit vi mi m khc 1. b) Xc nh gi tr ca m d phng tr nh c tch hai nghim bng 5, t hy tnh

tng hai nghim ca phng tr nh.c) Tm mt h thc lin h gia hai nghim khng ph thuc vo m.

d) Tm m phng tr nh c nghim 21; x x tho mn h thc 025

1

2

2

1

x x

x x

Bi 13 : Cho phng tr nh 012 mmx x a) Chng t rng phnh tr nh c nghim 21; x x vi mi m ; tnh nghim kp (nu c)

ca phng tr nh v gi tr ca m tng ng.

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b) t 212221 6 x x x x A i) Chng minh 882 mm A ii) Tm m A = 8iii) Tm gi tr nh nht ca A v gi tr ca m tng ng

c) Tm m sao cho phng tr nh c nghim ny bng hai ln nghim kia.

Bi 14: Cho phng tr nh 01222 mmx x a) Chng t rng phnh tr nh c nghim 21; x x vi mi m. b) t A = 212221 5)(2 x x x x

i) Chng minhA = 9188 2 mm ii) Tm m sao cho A = 27

c) Tm m sao cho phng tr nh c nghimny bng hai ln nghim kia.

Bi 15 : Gi s phng tr nh 0. 2 cbx xa c 2 nghim phn bit 21; x x . tnn

n x xS 21 vin l s nguyn dng. a) Chng minh 0.

12

nnncS bS S a

b) p dng tnh gi tr caA =55

251

251

Bi 16: Cho 2( ) 2( 2) 6 1 f x x m x m a) Chng minh phng tr nh ( ) 0 f x c nghim vi mi m. b) t 2t , tnh ( ) f x theo t, t tmiu kin i vi m phng tr nh

( ) 0 f x c 2 nghim ln hn 2.

Bi 17 : Cho phng tr nh 05412 22 mm xm x

a)

Xc nh gi tr ca m phng tr nh c nghim. b) Xc nh gi tr ca m phng tr nh c hai nghim phn bit u dng.c) Xc nh m phng tr nh c hai nghim c gi tr tuyt i bng nhau v tri du

nhau.d) Gi 21; x x l hai nghim nu c ca phng tr nh. Tnh 2221 x x theo m.

Bi 18: Cho phng tr nh 08342 x x c hai nghim l 21; x x . Khng gii phng

trnh, hy tnh gi tr ca biu thc2

31

321

2221

21

556106

x x x x x x x x

M

Bi 19 : Cho phng tr nh 2 2( 2) 1 0. x m x m

a) Gii phng tr nh khi 1 .2

m

b) Tm cc gi tr ca m phng tr nh c hai nghim tri duc) Gi 21; x x l hai nghim ca phng tr nh. Tm gi tr ca m

21221 )21()21( m x x x x

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Bi 20: Cho phng tr nh 032 nmx x (i)a) Cho n = 0, chng minh phng tr nh lun c nghim vi mi m.

b) Tm m v n hai nghim 21; x x ca phng tr nh (i) tho mn 71

22

21

21

x x

x x

Bi 21 : Cho phng tr nh 052222 k xk x a) Chng minh phngtrnh c hai nghim phn bit vi mi gi tr ca k b) Gi 21; x x l hai nghim ca phng tr nh. Tm gi tr ca k sao cho 182221 x x

Bi 22 : Cho phng tr nh 04412 2 mx xm a) Gii phng tr nh khi m = 1. b) Gii phng tr nh khi m ty .c) Tm gi tr ca m phng tr nh c mt nghim bng m.

Bi 23 : Cho phng tr nh 0332 22 mm xm x a) Chng minh phng tr nh lun c hai nghim phn bit vi mi m. b) Xc nh m phng tr nh c hai nghim 21, x x tho mn 61 21 x x

VN 3. H PHNG TR NH I S

A. PHNG PHP GII TON

Bi ton 3.1 Gii h phng tr nh sau10 5 1

12 3 4 17 8 1.12 3 4 1

x y

x y

Hng dn.K 1 1,4 4

x y , t 1 1,12 3 4 1

a b x y

vi , 0.a b

Khi , ta c h phng tr nh mi10 5 17 8 1

a b

a b

n y cc em lm tip, ch i chiu vi K khi tm ra k t qu. Bi ton 3.2 Gii h phng tr nh sau

1 1 4

(1 4 ) 2.

x y

x y y

(Trch thi tuyn sinh vo l p 10 tnh Nam nh nm 2011)

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Li gii.K , 0 x y , khi 1 1 4 4 y xy x y

Do (1 4 ) 2 4 2 2 x y y x xy y x x y y 2( ) 2 1 y x y

M1

4 4 1 4 xy x y xy xy . Nh vy1

1 ; .4 x y xy Do x, y l nghim ca PT

22 1 1 1 10 0 0

4 2 2 2

t t t t t

T 12

x y (tha mn K).

Vy 1 1; ;2 2

x y l nghim duy nht ca HPT cho.

Bi ton 3.3 Gii h phng tr nh sau3 2 17 (1)

2 1 52 2 2 26. (2)

2 1 5

x y

x y x y

( chung Chuyn L H ng Phong Nam nh nm 2011) Hng dn. K 2, 1, 1. x y y Khi (2) tng ng vi

2( 2) 2 2 26 2 2 2622 1 5 2 1 5

x y y

x y x y

2 2 16 6 3( 2) 482 1 5 2 1 5

y y

x y x y (i)

Vi 2, 1, 1 y y th 6 4 34 6 34 4(1)2 1 5 2 5 1

y x y

(ii)

T (i) v (ii) ta c: 34 4 3( 2) 48 3( 2) 4 145 1 1 5 1 1 5

y y y y y y

n y, cc em rt gn quy v phng tr nh bc hai v gii bnh thng. Bi ton 3.4 Gii h phng tr nh sau

2

2

1 31 3 .

x x y

y y x

Li gii.Tr v i v hai PT ta c 2 2 2 21 1 3 3 4 4 0 x x y y y x x y x y

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( )( ) 4( ) 0 ( )( 4) 0 x y x y x y x y x y

04 0 4

x y x y

x y y x

+ Vi x y th vo 2 1 3 x x y ta c 2 2 21 3 2 1 0 ( 1) 0 1 0 1 x x x x x x x

Do ( ; ) (1;1) x y l mt nghim ca HPT cho.+ Vi 4 x th vo 2 1 3 x x y ta c

2 2 21 3( 4) 4 13 0 ( 2) 9 0 x x x x x x (*)Mt khc 2 2( 2) 0 ( 2) 9 9 0 x x , do (*) v nghim. Vy ( ; ) (1;1) x y l nghim duy nht ca HPT cho.

Nh n xt. Khi ta thay i v tr ca x v y cho nhau th HPT khng thayi. Vi nhngHPT i xng nh trn, th ta s tr v cc PT vi nhau (thng th ta s thu c x = y,s dng kt qu ny phn tch thnh nhn t), sau th vo mt trong hai PT ca hr i gii PT mt n. Ta d dng chng minh c x v y dng bng cch lm sau y:

2 22 21 3 1 31 ; 1 .

2 4 2 4 x x x y y y

Bin 2 y y ta c HPT kh hnmt cht2

2

1 64 2 1 3 .

x y

y y x

i khi ngi ta li cho HPT gn i xng, chng hn ta xt bi ton sau.Bi ton 3.5 Gii h phng tr nh sau

2

2

1 21 3 .

x y

y y x

Hng dn.Tr v i v hai PT ta c 2 2 2 21 1 2 3 3 3 0 x y y y x x y x y

n y cc em gii nh bi ton trn. Bi ton 3.6 Gii h phng tr nh sau

2 2

2 2

3 52 2 4 4. x xy y

x xy y

Li gii.HPT cho tng ng vi

2 2

2 2

4 3 20

5 2 2 4 20

x xy y

x xy y

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2 2 2 22 2

2 2

2 2

4 3 5 2 2 4

6 16 22 03 8 11 03 3 8 8 03 ( ) 8 ( ) 0( )(3 8 ) 0

03 8 0 3 / 8

xy y x xy y

x y xy

x y xy

x xy y xy

x x y y x y

x y x y

x y x y

x y y x

+ Vi x = y, th vo HPT cho ta c2 2 2 2

22 2 2 2

3 5 5 51 1

2 2 4 4 4 4 x x x x

x x x x x x

Ta c 1 1, 1 1 ( ; ) (1;1), ( 1; 1) x y x y x y l 2 nghim ca HPT.+ Vi 3 / 8 x , cc em lm tng t nh trn.

Nh n xt. gii bi ton trn ta c th lm nh sau

+ Xt2

2

50

2 4 x

y x

HPT ny v nghim nn y = 0 khng tha mn.

+ Xt 0 y , t yt th vo HPT cho tac

2 22 2 2

2 2 2 2 2

3 1 53 . 52 2 . 4 4 2 2 4 4

y t t y t yt y y

y t yt y y y t t

V y khc 0 nn ta c

2 2 2

22 2

3 1 5 3 1 54 2 2 4 42 2 4

y t t t t t t y t t

n y cc em tmct suy ra mi lin h gia x v y r i gii nh trn.


Bi 1: Tm gi tr cam hc nghim duy nht tho mn iu kin y

nh nht

21

11 ym x

m y xm

Bi 2: Xc nh a v b h phng tr nh sauc v s nghim

542

aybx

by x

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Bi 3: Gii h phng tr nh sau trn R

11922

y xy x

y xy x

Bi 4: Tm m sao cho h phng tr nh sau c nghim

01121

2 y x y xm y x

y x


6241332

22

22

y xy x

y xy x

Bi 6: Tnh 22 ba bit rnga v b tho mn h phng tr nh

020342

222

23

bbaa

bba


2 2

34 6.

x y xy

x xy y

Bi 8: Gii h phng tr nh sau trn R3 3

2 2

34.

x y xy

x y x y

Bi 9: Gii h phng tr nh sau trn R2 2

2 2

5 2 43 2 3 2. x xy y

x xy y

Bi 10: Gii h phng tr nh sau

1 1 11

3 1 . x y

xy

(Trch thi tuy n sinh v o l p 10 tnh Nam nh nm 2012)

Bi 11: Cho h phng tr nha y xa

y xa.

3)1(

a) Gii h phng r nh khi 2a b) Xc nh gi tr ca a h c nghim duy nht tho mn 0. x y

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VN 4. CC BI TON V TH H M S

A. PHNG PHP GII TON

Xt parabol 2( ) : P y ax v ng thng( ) :d y mx n

Honh giao im ca (P) v (d) l nghim ca phng tr nh2 0 (*)ax mx n (C n lu thut ng ny trong gi i ton) (d) ct (P)ti 2 im phn bit khi v ch khi (*) c hai nghim phn bit. (d) ct (P) khi v ch khi (*) c nghim. (d) tip xc vi (P) khi v ch khi (*) c nghim kp. Ngoi ra cc em c n ch n bi ton tm m hai ng thng song song

vi nhau, vung gc vi nhau, hm s ng bin, nghch bin.


Bi 1: Cho hm s ( 2) ( ).m x n d Tm gi tr cam vn th (d ) ca hm s.a) i qua hai im ( 1;2), (3; 4). A B b) Ct trc tung ti im c tung bng1 2 v ct trc honh ti im c

honh bng2 2. c) Ct ng thng 2 3 0. x y d) Song song vii ng thng3 2 1. x y

Bi 2: Cho hm s 22 x y ( P ).a) V th ( P ). b) Tm trn th cc im cch u hai trc to .

c) Xt s giao im ca ( P ) vi ng thng (d ) : 1mx y theom.d) Vit phng tr nhng thng (d') i qua im M(0;-2) v tip xc vi (P).

Bi 3: Cho (P) : 2 x y v ng thng (d) : m x y 2 a) Xc nh m hai ng

i) Tip xc nhau. Tm to tip im.ii) Ct nhau ti hai im phn bit A v B, mt im c honh 1. x

Tm honh im cn li. Tm to A v B. b) Trong trng hp tng qut, gi s (d) ct (P) ti hai im phn bit M v N. Tm

to trung im I ca on MN theo m v tm qu tch ca im I khi m thay i. Bi 4: Cho ng thng (d) : 2)2()1(2 ym xm

a) Tm m ng thng (d) ct (P) : 2 x y ti hai im phn bit A v B. b) Tm to trung im I ca on AB theo m.c) Tm m (d) cch gc to mt khong Max.d) Tmim c nh m (d) i qua khi m thay i.

Bi 5: Cho (P) : 2 x y a) Tm tp hp cc im M sao cho t c th k c hai ng thng vung

gc vi nhau v tip xc vi (P).

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b) Tm trn (P) ccim sao cho khong cch ti gc to bng2.

Bi 6: Cho ng thng (d) : 343

x y

a) V (d). b) Tnh din tch tam gic c to thnh gia (d) v hai tr c to .c) Tnh khong cch t gc O n (d).

Bi 7: Cho hm s 1 x y (d)a) Nhn xt dng ca th. V th (d). b) Dng th bin lun s nghim ca phng tr nh m x 1

Bi 8: Vi gi trno ca m th haing thng(d) : 2)1( xm y ; (d) : 13 x y a) Song song vi nhau. b) Ct nhau. c) Vung gc vi nhau.

Bi 9: Tm gi tr ca a ba ng thng sau ng quy ti mt im trong mt phngto .

1 2 3: 2 5 ; : 2 ; : 12.d y x d y x d y ax Bi 10: Chng minh rng khi m thay i th ( ) : 2 ( 1) 1d x m y lun i qua mt imc nh.

Bi 11: Cho (P) : 221

x y v ng thng( ) : .d y ax b Xc nh a v b ng

thng (d) i qua im A(-1;0) v tip xc vi (P).

Bi 12: Cho hm s 21 x x y a) V th hn s tr n. b) Dng th cu a bin lun theo m s nghim ca phng tr nh

m x x 21 Bi 13: Cho 2( ) : ; ( ) : 2 . P y x d y x m

a) V (P). b) Tm m (P) tip xc (d).

Bi 14: Cho2

( ) : ; ( ) : .4

x P y d y x m

a) V (P). b) Xc nh m (P) v (d) ct nhau ti hai im phn bit A v B.c) Xc nh phng tr nhng thng (d') song song vi ng thng (d) v ct (P) ti

im c tung bng-4.d) Xc nh phng tr nhng thng (d'') vung gc vi (d') v i qua giao im ca

(d') v (P)

Bi 15: Cho hm s 2 x y (P) v hm s y = x + m (d)a) Tm m sao cho (P) v (d) ct nhau ti hai im phn bit A v B b) Xc nh phng tr nhng thng (d') vung gc vi (d) v tip xc vi (P)

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c) Thit lp cng thc tnh khongcch gia hai im bt k . p dng tm m sao chokhong cch gia hai im A v B bng 23

Bi 16: Cho im A(-2;2) v ng thng 1 : 2( 1).d y x a) im A c thuc 1d khng. b) Tm a hm s 2. xa y (P) i qua A.c) Xc nh phng tr nhng thng 2d i qua A v vung gc vi 1 .d d) Gi A v B l giao im ca (P) v 2d ; C l giao im ca 1d vi trc tung.

Tm to ca B v C. Tnh din tch tam gic ABC.

Bi 17: Cho (P) : 241

x y v ng thng (d) qua hai im A v B trn (P) c honh

lm lt l -2 v 4.a) Kho st s bin thin v v th (P) ca hm s tr n. b) Vit phng tr nhng thng (d).

c)

Tm im M tr n cung AB ca (P) tng ng honh 4;2 x sao cho tamgic MAB c din tch ln nht. Bi 18: Cho (P) :

4

2 x y v im M (1;-2).

a) Vit phng tr nhng thng (d) i qua M v c h s gc l m. b) Chng minh-+(d) lun ct (P) ti hai im phn bit A v B khi m thay i.c) Gi B A x x ; ln lt l honh ca A v B.Xc nh m 22 B A B A x x x x t gi tr

nh nht v tnh gi tr .d) Gi A' v B' ln lt l hnh chiu ca A v B trn tr c honh v S l din tch t

gic AA'B'B. Tnh S theo m.

Bi 19: Cho hm s2

x y

(P)a) V (P). b) Gi A, B l hai im thuc (P) c honh ln lt l -1 v 2. Vit phng

trnhng thng AB.c) Vit phng tr nhng thng (d) song song vi AB v tip xc vi (P).

Bi 20: Cho parabol (P) : 241

x y v ng thng (d): 12 mmx y

a) V (P). b) Tm m sao cho (P) v (d) tip xc nhau. Tm to tip im.c) Chng t rng (d) lun i qua mt imc nh.

Bi 21: Cho (P) : 241 x y v im I(0;-2). Gi (d) l ng thng qua I v c h s

gc m.a) V (P) v chng minh(d) lun ct (P) ti hai im phn bit A v B. b) Tm gi tr ca m on AB ngn nht.

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Bi 22: Cho (P) :4

2 x y v ng thng (d) i qua im(3/ 2;1) I c h s gc l m.

a) V (P) v vit phng tr nh (d). b) Tm m sao cho (d) tip xc (P).c) Tm m sao cho (d) v (P) c haiim chung phn bit.

Bi 23: Cho (P) :42 x y v ng thng (d) : 2

2 x y

a) V (P) v (d). b) Tm to giao im ca (P) v (d).c) Tm to ca im thuc (P) sao cho ti ng tip tuyn ca (P) song song

vi (d).

Bi 24: Cho (P) : 2 x y a) V (P). b) Gi A v B l hai im thuc (P) c honh ln lt l -1 v 2. Vit phng tr nh

ng thng AB.c) Vit phng tr nhng thng (d) song song vi AB v tip xc vi (P).

Bi 25: Cho (P) : 22 x y a) V (P). b) Trn (P) ly im A c honh x = 1 v im B c honh x = 2. Xc nh cc

gi tr ca m v n ng thng (d) : y = mx + n tip xc vi (P) v song song viAB.

Bi 26: Xc nh gi tr ca m hai ng thng 1 2: ; : 1d x y m d mx y ctnhau ti mt im tr n 2( ) : 2 . P y x

VN 5. GII TON BNG CCH LP PHNG TR

NH

D NG 1. TON CHUYN NG

Bi 1: Hai tnh A v B cch nhau 180 km. Cng mt lc, mt t i t A n B v mt xe my i t B v A. Hai xe gp nhau ti th trn C. T C n B t i ht 2 gi , cn t Cv A xe my i ht 4 gi 30 pht. Tnh vn tc ca mi xe bit rng trn ng AB haixe u chy vi vn tc khng i. Bi 2: Mt ca n xuidng t bn A n bn B ri li ngc dng t bn B v bn Amt tt c 4 gi . Tnh vn tc ca ca n khi nc yn lng, bit rng qung sng AB di30 km v vn tc dng nc l 4 km/h.

Bi 3: Mt ca n xui t bn A n bn B vi vn tc 30 km/h, sau li ngc t B trv A.Thi gian xui t hn thi gian i ngc 1 gi 20 pht. Tnh khong cch gia hai bn A v B bit rng vn tc dng nc l 5 km/h.Bi 4: Mt ngi chuyn ng u tr n mt qungng gm mt on ng bng vmt on ng dc. Vn tc trn on ng bng v trn on ng dc tng ng l40 km/h v 20 km/h. Bit rng on ng dc ngn hn on ng bng l 110km vthi gian ngi ic qung ng l 3 gi 30 pht. Tnh chiu di qungngngi i.

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Bi 5: Mt xe ti v mt xe con cng khi hnh t A n B. Xe ti i vi vn tc 30

km/h, xe con i vi vn tc 45 km/h. Sau khi i c43 qungng AB, xe con tng

vn tc thm 5 km/h trn qungng cn li. Tnh qungng AB bit rng xe conn B sm hn xe ti 2gi 20 pht. Bi 6: Mt ngi i xe p t A n Bcch nhau 33 Km vi mt vn tc xc nh Khit B v A ngi i bng con ng khc di hn trc 29 Km nhng vi vn tc lnhn vn tc lc i 3 Km/h. Tnh vn tc lc i , bit rng thi gian v nhiu hn thi giani l 1 gi 30 pht. Bi 7: Hai ca n cng khi hnh t hai bn A, B cch nhau 85 Km i ngc chiu nhau.Sau 1h40 th gp nhau. Tnh vn tc ring ca mi ca n, bit rng vn tc ca n i xuiln hn vn tc ca n i ngc 9Km/h v vn tc dng nc l 3 Km/h.

Bi 8: Hai a im A,B cch nhau 56 Km . Lc 6h45pht mt ngi i xe p t A vivn tc 10 Km/h. Sau 2 gi mt ngi i xe p t B n A vi vn tc 14 Km/h. Hi

n my gi h gp nhau v ch gp nhau cchA bao nhiu Km.Bi 9: Mt ngi i xe p t A n B vi vn tc 15 Km/h. Sau mt thi gian, mtngi i xe my cng xut pht t A vi vn tc 30 Km/h v nu khng c g thayi ths ui kp ngi i xe my ti B. Nhng sau khi i c na qung ng AB, ngi ixe p gim bt vn tc 3 Km/h nn hai ngi gp nhau ti C cch B 10 Km . Tnhqungng AB

Bi 10: Mt ngi i xe my t A n B vi vn tc trung bnh l 30 Km/h . Khin Bngi ngh 20 pht ri quay tr v A vi vn tc trung bnh l 24 Km/h . Tnh qungng AB bit rng thi gian c i ln v l 5 gi 50 pht.

Bi 11: Mt ca n xui t bn A n bn B vi vn tc trung bnh 30 Km/h , sau ngc t B v A. Thi gian i xui t hn thi gian i ngc l 40 pht. Tnh khongcch gia hai bn A v B bit rng vn tc dng nc l 3 Km/h v vn tc ring ca can l khng i .

Bi 12: Mt t d nh i t tnh A n tnh B vi vvn tc trung bnh l 40 Km/h . Lcu t i vi vn tc , khi cn 60 Km na th c mt na qung ng AB ,ngi li xe tng vn tc thm 10 Km/h trn qungng cn li . Do t n tnh Bsm hn 1gi so vi d nh . Tnh qungng AB.

Bi 13: Hai ca n khi hnh cng mt lc v chy t bn A n bn B . Ca n I chy vivn tc 20 Km/h , ca n II chy vi vn tc 24 Km/h . Trn ng i ca n II dng li 40 pht , sau tip tc chy . Tnh chiu di qungng sng AB bit rng hai ca n nB cng mt lc . Bi 14: Mt ngi i xe p t A n B cch nhau 50 Km . Sau 1 gi 30 pht , mtngi i xe my cng i t A v n B sm hn 1 gi . Tnh vn tc ca mi xe , bitr ng vn tc ca xe my gp 2,5 ln vn tc xe p.

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Bi 25: Hai t sn xut cng nhn chung mt mc khon . Nu lm chung trong 4 gi th

hon thnh c32 mc khon . Nu mi t lm ring th t ny s lm xong mc

khon th mi t phi lm trong bao lu ?

Bi 26: Hai t cng nhn lm chung trong 12 gi s hon thnh xong cng vic nh .H lm chung vi nhau trong 4 gi th t th nht c iu i lm vic khc , t th hailm nt cng vic cn li trong 10 gi . Hi t th hai lm mt mnh th sau bao lu shon thnh cng vic.

Bi 27: Hai ngi th cng lm mt cng vic trong 16 gi th xong . Nu ngi thnht lm 3 gi v ngi th hai lm 6 gi th h lm c 25% cngvic. Hi mi ngilm cng vic trong my gi th xong.

D NG 3. TON TH TCH

Bi 28: Hai vi n

c cng ch

y vo m

t ci b khng cha nc l

m y b trong 5gi 50 pht Nu chy ring th vi th hai chy y b nhanh hn vi th nht l 4 gi .Hi nu chy ring th mi vi chy trong bao lu s y b ?

Bi 29: Hai vi nc cng chy vo mt ci b khng c nc v chy y b mt 1 gi48 pht . Nu chy ring , vi th nht chy y b nhanh hn vi th hai trong 1 gi 30 pht . Hi nu chy ring th mi vi s chy y b trong bao lu ?

Bi 30: Mt my bm mun bm y nc vo mt b cha trong mt thi gian quy

nh th mi gi phi bm c 10 m3 . Sau khi bm c31 th tch b cha , my

bm hot ng vi cng sut ln hn , mi gi bm c 15 m3 . Do vy so vi quynh , b cha c bm y trc 48 pht. Tnh th tch b cha.

Bi 31: Nu hai vi nc cng chy vo mt ci b cha khng c nc th sau 1 gi 30 pht s y b . Nu m vi th nht trong 15 pht ri kho li v m vi th hai chy

tip trong 20 pht th s c51 b . Hi mi vi chy ring th sau bao lu s y b ?

Bi 32: Hai vi nc cng chy vo mt ci b cha khng c nc th sau 2 gi 55 phts y b Nu chy ring th vi th nht chy y b nhanh hn vi th hai 2 gi . Hinu chy ring th mi vi chy y b trong bao lu.

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VN 6. CC BI TON HNH H C TNG HP

Bi 1: Cho hai ng tr n tm O v O c R > R tip xc ngoi ti C.K cc ng knhCOA v COB. Qua trung im M ca AB, dng DE vung gc vi BC.

a) T gic ADBE l hnh g. b) Ni D vi C ct ng tr n tm O ti F. Chng minh B,E, F thng hng.c) Ni D vi B ct ng tr n tm O ti G. Chng minh EC i qua G.d) Xt v tr ca MF i vi ng tr n tm O, v tr ca AE vi ng tr n ngoi

tip t gic MCFE.Bi 2: Cho na ng tr n tm Ong knh CD = 2R. Dng Cx, Dy vung gc viCD. T im E bt k trn na ng tr n, dng tip tuyn vi ng tr n , ct Cx ti P ,ct Dy ti Q.

a) Chng minhtam gic POQ vung v POQ ng dng vi CED. b) Tnh tch CP.DQ theo R.c) Tnh th tch ca hnh gii hn bi na ng tr n tm O v hnh thang vung

CPQD khi chng cng quay theo mt chiu v tr n mt vng quanh CD.Bi 3: Cho ng tr n tm O bn knh R c haing knh AOB , COD vung gc vinhau. Ly im E bt k trn OA , ni CE ct ng tr n ti F . Qua F dng tip tuyn Fxvi ng tr n , qua E dng Ey vung gc vi OA . Gi I l giao im ca Fx v Ey .

a) Chng minh I, F, E, O cng thuc mt ng tr n. b) T gic CEIO l hnh g.c) Khi E chuyn ng tr n AB th I chuyn ng trn ng no.

Bi 4: Cho ng tr n tm O v mt im A trn ng tr n . Qua A dng tip tuynAx . Trn Ax ly mt im Q bt k , dng tip tuyn QB .

a) Chng minh t gic QBOA ni tip. b) Gi E l trung im ca QO , tm qu tch ca E khi Q chuyn ng tr n Ax.c) H BK Ax , BK ct QO ti H . Chng minh t gic OBHA l hnh thoi, t

suy ra qu tch ca im H.Bi 5: Cho ABC c ba gc nhn ni tip ng tr n tm O. Cc ng cao AD , BKct nhau ti H , BK ko di ct ng trong ti F . V ng knh BOE.

a) T gic AFEC l hnh g. b) Gi I l trung im ca AC. Chng minh H, I, E thng hng.c) Chng minhOI =

2 BH v H, F i xng nhau qua AC.

Bi 6: Cho (O, R) v (O, R ) vi R > R tip xc trong ti A . ng ni tm ct ngtrn O v ng tr n O ti B v C . Qua trung im P ca BC dng dy MN vung gcvi BC . Ni A vi M ct ng tr n O ti E .

a) So snh hai gc AMO v NMC. b) Chng minh N , B , E thng hng v OP = R ; OP = R.c) Xt v tr ca PE vi ng tr n tm O.

Bi 7: Cho ng tr n tm Ong knh AB . Ly B lm tm v ng tr n bn knhOB . ng tr n ny ct ng tr n O ti C v D

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a) T gic ODBC l hnh g. b) Chng minh OC AD ; OD ACc) Chng minh tr c tm ca tam gic CDB nm trn ng tr n tm B.

Bi 8: Cho ng tr n tm O v mt ng thng d ct ng tr n ti hai im cnh A v B . T mt im M bt k trnng thng d nm ngoi on AB ngi ta khai tip tuyn vi ng tr n l MP v MQ ( P, Q l cc tip im ).

a) Tnh cc gc ca MPQ bit rng gc gia hai tip tuyn MP v MQ l 450 b) Gi I l trung im AB . Chng minhM , P , Q , O , I cng nm tr n mt ng

trn.c) Tm qu tch tm ng tr n ngoi tip MPQ khi M chy tr n d.

Bi 9: Cho ABC ni tip ng tr n tm O , tia phn gic trong ca gc A ct cnh BCti E v ct ng tr n ti M .

a) Chng minh OM BC b) Dng tia phn gic ngoi Ax ca gc A . Chng minh Ax i qua mt im c nhc) Ko di Ax ct CB ko di ti F . Chng minh:FB . EC = FC . EB

Bi 10: Cho ABC c AB = AC v gc BAC nhn, mt cung tr n BC nm trong ABC v tip xc vi AB , AC ti B v C . Trn cung BC ly im M ri h cc ngvung gc MI , MH , MK xung cc cnh tng ng BC , CA , AB . Gi P l giao imca MB , IK v Q l giao im ca MC , IH.

a) CMR cc t gic BIMK , CIMH ni tip b) CMR tia i ca tia MI l phn gic ca gcHMKc) CMR t gic MPIQ ni tip c, t suy raPQ BC

Bi 11: : Cho ABC c AC > AB v gc BAC t. GiI , K theo th t l cc trung imca AB , AC . Cc ng tr n ng knh AB , AC ct nhau ti im th hai D ; tia BAct ng tr n (K) ti im th hai E ; tia CA ct ng tr n (I) ti im th hai F.

a) CMR ba im B , C , D thng hng b) CMR t gic BFEC ni tip cc) Chng minh ba ng thng AD , BF , CE ng quy d) Gi H l giao im th hai ca tia DF vi ng tr n ngoi tip AEF . Hy so

snh di cc on thng DH , DE .

Bi 12: Cho ng tr n (O;R) vim A vi OA = 2 R , mt ng thng (d) quayquanh A ct (O) ti M , N ; gi I l trung im ca on MN.

a) Chng minhOI MN. Suy ra I di chuyn tr n mt cung tr n c nh vi haiim gii hn B , C thuc (O).

b) Tnh theo R di AB , AC. Suy ra A, O, B, C l bn nh ca hnh vung.

c) Tnh din tch ca phn mt phng gii hn bi on AB, AC v cung nh BCca (O).Bi 13: Cho na ng tr n ng knh AB = 2R , C l trung im ca cung AB . Tr ncung AC ly im F bt k . Trn dy BF ly im E sao cho BE = AF.

a) AFC v BEC c quan hvi nhau nh th no. b) CMR FEC vung cnc) Gi D l giao im ca ng thng AC vi tip tuyn ti B ca na ng tr n .

CMR t gic BECD ni tip c.

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b) Chng minh MN l tia phn gic ca gc AMB v ng thng MN lun i quamt im K c nh.

c) Chng minh KM.KN khng i.Bi 21: Cho mt ng tr n ng knh AB, cc im C, D trn ng trn sao cho C,D khng nm tr n cng mt na mt phng b AB ng thi AD > AC. Gi cc imchnh gia cc cung AC , AD ln lt l M , N ; giao im ca MN vi AC , AD ln ltl H , I ; giao im ca MD vi CN l K

a) Chng minh MAK NKD ; cn. b) Chng minht gic MCKH ni tip v KH // AD.c) So snh gc CAK vi gc DAK .

Bi 22: Cho ba im A , B , Ctrn mt ng thng theo th t y v ng thng (d)vung gc vi AC ti A . V ng tr n ng knh BC v trn ly im M bt k .Tia CM ct ng thng d ti D ; tia AM ct ng tr n ti im th hai N ; tia DB ctng tr n ti im th hai P.

a) Chng minh t gic ABMD ni tip.

b)

Chng minhCM.CD khng ph thuc v tr ca Mc) T gic APND l hnh g.d) Chng minh trng tm G ca tam gic MAC chy tr n mt ng tr n c nh khi

M di ng. Bi 23: Cho na ng tr n tm Ong knh AB . Mt im Mnm tr n cung AB ;gi H l im chnh gia ca cung AM . Tia BH ct AM ti mt im I v ct tip tuynti A ca ng tr n (O) ti im K . Cc tia AH ; BM ct nhau ti S .

a) Tam gic BAS l tam gic g ? Ti sao ? Suy ra im S nm tr n mt ng tr nc nh .

b) Xc nh v tr tong i ca ng thng KS vi ng tr n (B;BA).c) ng tr n i qua B, I, S ct ng tr n (B;BA) ti mt im N. Chng minh

ng thng MN lun i qua mt im c nh khi M di ng tr n cung AB.d) Xc nh v tr ca M sao chogc MKA bng 90 . Bi 24 : Cho t gic ABCD ni tip trong mt ng tr n v P lim chnh gia cacung AB khng cha C v D . Hai dy PC v PD ln lt ct dy AB ti E v F . Cc dyAD v PC ko di ct nhau ti I ; cc dy BC v PD ko di ct nhau ti K. Chng minhr ng

a) Gc CID bng gc CKD. b) T gic CDFE ni tip c.c) IK // AB.d) ng tr n ngoi tip tam gic AFD tip xc vi PA ti A.

Bi 25: Cho hai ng tr n (O1) v (O2) tip xc ngoi vi nhau ti A , k tip tuynchung Ax. Mt ng thng d tip xc vi (O1) , (O2) ln lt ti cc im B , C v ctAx ti im M . K cc ng knh BO1D v CO2E.

a) Chng minh M l trung im ca BC b) Chng minh tam gicO1MO2 vungc) Chng minh B , A , E thng hng v C , A , D thng hngd) Gi I l trung im caDE. Chng minh ng tr n ngoi tip tam gic IO1O2

tip xc vi ng thng d.

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Bi 26: Cho (O;R) trn c mt dy AB = R 2 c nh v mt im M di ng tr ncung ln AB sao cho tam gic MAB c ba gc nhn . Gi H l tr c tm ca tam gicMAB ; P , Q ln lt l cc giao im th hai ca cc ng thng AH , BH vi ngtrn (O) ; S l giaoim ca cc ng thng PB , QA.

a) Chng minh PQ l ng knh ca ng tr n (O)

b) T gic AMBS l hnh gc) Chng minh di SH khng id) Gi I l giao im ca cc ng thng SH, PQ. Chng minh I chy tr n mt

ng tr n c nh. Bi 27: Cho (O;R) ng knh AB, k ti p tuyn Ax v trn ly imP sao cho AP >R . K tip tuyn PM (M l tip im ).

a) Chng minh BM // OP b) ngthng vung gcvi AB ti O ct tiaBM ti N. T gic OBNP l hnh g.c) Gi K l giao im ca AN vi OP ; I l giao im ca ON vi PM ; J l giao

im ca PN vi OM. Chng minh K, I, J thng hngd) Xc nh v tr ca P sao cho K nm trn ng tr n (O).

Bi 28: Cho ng tr n (O;R) , haing knh AB v CD vung gc nhau . Trong onthng AB ly im M ( khc im O ) , ng thng CM ct ng tr n (O) ti im thhai N . ng thng vung gc vi AB ti M ct tip tuyn ti N vi ng tr n (O)im P .

a) Chng minh t gic OMNP ni ti p. b) T gic CMPO l hnh g.c) Chng minh CM.CN khng i.d) Chng minh khi M di ng trn on AB th P chy tr n mtng thng c

nh.Bi 29: Cho hai ng tr n (O), (O) ct nhau ti hai im A v B. Cc ng thng AO,AO ct ng tr n (O) ln lt ti cc im th hai C, D v ct ng tr n (O) ln ltti cc im th hai E, F.

a) Chng minh B, F, C thng hng. b) Chng minh t gic CDEF ni tip. c) Chng minh A l tm ng tr n ni tip tam gic BDE.d) Tmiu kin DE l tip tuyn chung ca cc ng tr n (O) v (O).

Bi 30: Cho na ng tr n ng knh AB = 2R v mt im M bt k tr n na ngtrn (M khc A v B). ng thng d tip xc vi na ng tr n ti M v ct ngtrung tr c ca on ABti I. ng tr n (I) tip xc vi AB ct ng thng d ti C vD (D nm trong gc BOM).

a) Chng minh cc tia OC, OD l cc tia phn gic ca cc gc AOM, BOM. b) Chng minhCA v DB vung gc vi AB.c) Chng minh tam gic AMB v COD ng dng. d) Chng minh h thc: AC.BD = R 2.

Bi 31: Cho ng tr n (O;R) ng knh AB v mt im M bt k trn ng tr n.Gi cc im chnh gia ca cc cung AM , MB ln lt l H, I. Cc dy AM v HI ctnhau ti K.

a) Chng minh gc HKM c ln khng i.

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b) H . Chng minh IP l tip tuyn ca (O;R).c) Gi Q l trung im ca dy MB. V hnh bnh hnh APQS. Chng minh Sthuc ng tr n (O;R).d) Chng minh khi M di ng th th ng thng HI lun lun tip xc vi mtng tr n c nh.

Bi 32: Cho na ng tr n (O)ng knh AB v hai im C , D thuc na ng tr nsao cho cung AC < 900 v gc COD bng 90 . Gi M l mt im tr n na ng tr nsao cho C l im chnh chnh gia cung AM. Cc dy AM, BM ct OC, OD ln lt tiE v F.

a) T gic OEMF l hnh g. b) Chng minh D l im chnh gia ca cung MB. c) Mt ng thng d tip xc vi na ng trn ti M v ct cc tia OC , OD

ln lt ti I v K. Chng minh cc t gic OBKM, OAIM ni tip.d) Gi s tia AM ct tia BD ti S. Xc nh v tr ca C v D sao cho 5 im M , O

B , K , S cng thuc mt ng tr n.Bi 33: Cho tam gic ABC c AB = AC , mt cung tr n BC nm bn trong tam gicABC v tip xc vi AB , AC ti B , C sao cho A v tm ca cung BC nm khc pha ivi BC . Tr n cung BC ly mt im M ri k cc ng vung gc MI , MH , MKxung cc cnh tng ng BC , CA , AB. Gi giao im ca BM , IK l P ; giao imca CM , IH l Q. Chng minh rng

a) Cc t gic BIMK, CIMH ni tip c. b) MI2 = MH . MKc) T gic IPMQ ni tip v PQ MId) Nu KI = KB th IH =

VN 7. MT S TON LUYN THI

thi th s 1. Thi gian 120 pht

Cu I

Cho biu thc P =

a

a

a

aa

a

a

a11.

1112 3

3 3

1) Rt gn P 2) Xt du ca biu thc 1 . P a

Cu II Mt ca n xui t A n B vi vn tc 30km/h, sau li ngc t B v A. Thigian xui t hn thi gian ngc 1h20 pht. Tnh khong cch gia hai bn A v B bit rng vn tc dng nuc l 5km/h v vn tc ring ca ca n khi xui v ngcl bng nhau.

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Cu III

Cho tam gic ABC cn ti A vi gc BAC nhn. Mt cung tr n BC nm trong tamgic ABC v tip xc vi AB, AC ti B v C. Trn cung BC ly mt im M ri hng vung gcMI, MH, MK xung cc cnh tng ng BC, AB, CA. Gi P lgiao im ca MB, IK v Q l giao im ca MC, IH.1) Chng minh rng cc t gic BIMK, CIMH ni tip c.2) Chng minh tia ica tia MI l phn gic ca gc HMK .3) Chng minh tgic MPIQ ni tip v PQ song song viBC.4) Gi (O2) l ng tr n i qua M, P, K ; (O2) l ng tr n i qua M, Q, H ; N l

giao im th hai ca (O1) v (O2) v D l trung im ca BC. Chng minh M, N,D thng hng.

Cu IVTm tt c cc cp s ( x, y) tho mn phng tr nh 25 2 ( 2) 1 0. x x y y

thi th s 2. Thi gian 120 pht

Cu I

Cho biu thc A =

12

21:1

11

a

a

a

a

aa

1) Rt gn A 2) Tm a 1/6. A

Cu IICho phngtrnh 2 2( 2) 1 0 x m x m vi m l tham s. 1) Gii phng tr nh khi 3/2.m 2) Tm m phung tr nh c hai nghim tri du.3) Gi 1 2, x x l hai nghim ca phng tr nh .Tm m

21 2 2 11 2 1 2 . x x x x m Cu III

Cho tam gic ABC( , 90 ). AB AC BAC Gi I, K th t l cc trung im caAB vAC. Cc ng tr n ng knh AB, AC ct nhau ti im th hai D; tia BAct ng tr n (K) ti im th hai E, tia CA ct ng tr n (I) ti im th hai F. a) Chng minh bai im B, C, D thng hng. b) Chng minh t gic BFEC ni tip. c) Chng minh ba ng thng AD, BF, CE ng quy.

d)

Gi H l giao im th hai ca tia DF vi ng tr n ngoi tip tam gic AEF.Hy so snh di cc on thng DH, DE.Cu IV

Xt hai phng tr nh bc hai 2 20 v 0.ax bx c cx bx a Tm h thc giaa,b, c l iu kin cn v hai phng tr nh trn c mt nghim chung duy nht.

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thi th s 3. Th i gian 120 pht

Cu I

Cho biu thc A =

12

11:

122

11

x x x x x x

x

x

1) Rt gn A 2) V i GT no ca x th At GTNN v tm GTNN.

Cu II

Mt ngi i xe my t A n B cch nhau 120km vivn tc d nh trc .Saukhi i c1/3 qung ng AB ngi tng vn tc ln 10km/h trn qungng cn li. Tm vn tc d nh v thi gian ln bnh trn ng,bit rngngi n B sm hn d nh 24pht.

Cu III

Cho ng tr n (O) bn knh R v mt dy BC c nh. Gi A l im chnh gia cacung nh BC. Ly im M tr n cung nh AC, k tia Bx vung gc vi tia MA I vct tia CM ti D. 1) Chng minhgc AMD = gc ABC v MA l tia phn gic ca gc BMD. 2) Chng minh A l tm ng tr n ngoi tip tam gic BCD v gc BDC c ln

khng ph thuc vo v tr im M. 3) Tia DA ct tia BC ti E v ct ng tr n (O) ti im th hai F, chngminh AB

l tip tuyn ca ng tr n ngoai tip tam gic BEF. 4) Chng minh tch P = AE.AF khng i khi M di ng. Tnh P theo bn knh R v

ABC = Cu IV

Cho hai bt phng tr nh 3 2 1 v 2 0.mx m x m x Tm m hai bt phngtrnh trn c cng tp hp nghim.


Cu I

Cho biu thc P =

141:

11

112

3 x x

x

x x

x

1) Rt gn P 2) Tm gi tr nguyn ca x P nhn gi tr nguyn dng.

Cu II

Mt ngi d nh i xe p t A n B cch nhau 96km trong thi gian nhtnh.Sau khi i c na qungng ngi dng li ngh 18 pht.Do n B ng hn ngi tng vn tc thm 2km/h trn qungng cn li.Tnh vn tc ban u v thi gian xe ln bnh trn ng.

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Cu III

Cho tam gic ABC vung ti A, ng cao AH. ng tr n ng knh AH ctcc cnh AB, AC ln lt ti E v F.1) Chng minh t gic AEHF l hnh ch nht.2) Chng minhAE.AB = AF.AC3) ng thng qua A vung gc vi EF ct cnh BC ti I. CMR: I l trungim ca BC.4) Chng minhnu din tch tam giac ABC gp i din tch hnh ch nht

AEHF th tam gic ABC vung cn.


Cu I

Cho biu thc P =

12

11:1

1 x x x x x x

1) Rt gn P 2) Tm x 0. P 3) Tm m . P x m x

Cu IIMt xe ti v mt xe con cng khi hnh t A i n B. Xe ti i vi vn tc40km/h, xe con i vi vn tc 60km/h. Saukhi mi xe i c na ng th xecon ngh 40 pht ri chy tip n B; xe ti tr n qungng cn li tng vntc thm 10km/h nhng vn n B chm hn xe con na gi. Hy tnh qung-ng AB.

Cu III

Cho ng tr n (O) v mt im A nm ngoi ng tr n. T A k hai tip tuynAB, AC v ct tuyn AMN vi ng tr n (B, C, M, N thuc ng tr n, AM

On Thi Lop 10 Toan VuVanBac

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