Các PP giai Toan lop 9

7/28/2019 Cc PP giai Toan lop 9

1/63

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LUYN THI VO LP 10

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Chuyn 1:

Bin i ng thc - Phn tch a thc thnh nhn t

A. bin i ng thc

I. Cc hng ng thc c bn v m rng

(a b)2 = a2 2ab + b2

a2 - b2 = (a + b)(a - b)

(a b)3 = a3 3a2b + 3ab2 b3

a3 - b3 = (a - b)(a2 + ab + b2)

a3 + b3 = (a + b)(a2 - ab +b2)

(a + b + c)2 = a2 + b2 + c2 + 2ab + 2ac + 2bc

(a - b - c)2 = a2 + b2 + c2 - 2ab - 2ac + 2bc

an - bn = (a - b)(an-1 + an-2b + ... + abn-2 + bn-1), mi n l s t nhin

an + bn = (a + b)(an-1 - an-2b + ... - abn-2 + bn-1), mi n l

II. Bi tp

Bi 1

So snh hai s A v B bit: A = 2004.2006 v B = 20052

Gii

Ta c A = (2005 - 1)(2005 + 1) = 20052 - 1 < 20052 =B. Vy A < B.

Bi 2

So snh hai s A v B bit: A = (2 + 1)(22 +1)(24 + 1)(28 + 1)(216 + 1) v B = 232

Gii

Ta c A = (2 - 1)(2 + 1)(22 +1)(24 + 1)(28 + 1)(216 + 1) = 232 -1 < 232 = B. Vy A b > 0, tho mn: 3a2 + 3b2 =10ab. Chng minh rng: =+

a b 1a b 2

.

Gii

t P =ba

ba

+

th P > 0 nn P = 2P .

Ta c P2 = + + = = =+ + + +

2 2 2 2

2 2 2 2

a b 2ab 3a 3b 6ab 10ab 6ab 1a b 2ab 3a 3b 6ab 10ab 6ab 4

. Vy P = 1/2.



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Bi 10

Cho a + b + c = 1 v + + =1 1 1

0a b c

. Chng minh rng: a2 + b2 + c2 =1.

Gii

T: a + b + c = 1 a2 + b2 + c2 + 2(ab + ac + bc) = 1 a2 + b2 + c2 = 1- 2(ab +ac + bc) .

Mt khc:+ ++ + = = + + =1 1 1 ab ac bc0 0 ab ac bc 0

a b c abc. Vy: a2 + b2 + c2 =1.

Bi 11

Cho + + =1 1 1

2a b c

(1) v a + b + c = abc. Chng minh rng: + + =2 2 21 1 1

2a b c

Gii

(1)+ ++ + + + + = + + + =2 2 2 2 2 2

1 1 1 1 1 1 1 1 1 a b c2( ) 4 2( ) 4

a b c ab ac bc a b c abc.

Thay a + b + c = abc vo ta c + + + = + + =2 2 2 2 2 21 1 1 1 1 1

2 4 2a b c a b c

.

Bi 12

Cho + + =x y z

1a b c

(1), v + + =a b c

1x y z

(2). CMR: = + + =2 2 2

2 2 2

x y zA 1

a b c

Gii

+ ++ + + + + = = + + =

2 2 2

2 2 2

x y z xy xz yz xy xz yz cxy bxz ayz2( ) 1 A 1 2( ) 1 2( )

a b c ab ac bc ab ac bc abc

(2):+ +

=cxy bxz ayz

0xyz

. Vy A = 1.

Bi 13

Cho + + =1 1 1

0a b c

.(1) Chng minh rng: + + =3 3 31 1 1 3a b c abc

.

Gii .

(1) = + = + + + = + + 3 3 3 3 3 31 1 1 1 1 1 1 1 1 1 1 1 1 1

( ) ( 3 ( ) [ 3 ( )]a b c a b c bc b c a b c bc a

Vy + + =3 3 31 1 1 3a b c abc

.

Bi 14

Cho a + b + c = 0 v a2 + b2 + c2 =14. Chng minh rng: a4 + b4 + c4 = 98.

Gii

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T: a + b + c = 0 a = -(b + c) a2 = (b + c)2 a2 = b2 + c2 +2bc

a2 - b2 - c2 = 2bc (a2 - b2 - c2)2 = 4b2c2 a4 + b4 + c4 - 2a2b2 - 2a2c2 + 2b2c2 =

4b2c2 a4 + b4 + c4 = 2a2b2 + 2b2c2 + 2a2c2 2(a4 + b4 + c4) = a4 + b4 + c4 +

2a2b2 - 2b2c2 + 2a2c22(a4 + b4 + c4 ) = (a2 + b2 + c2 )2 = 142 =196.

Vy a4 + b4 + c4 = 98.

Bi 15

Cho xyz = 1, Chng minh rng: + + =+ + + + + +1 1 1

1.1 x xy 1 y yz 1 z zx

Gii

Ta c: + + = + + =+ + + + + + + + + + + +

1 1 1 z x 1

1 x xy 1 y yz 1 z zx z xz xyz x yx xyz 1 z zx

=+ +

+ + = + = ++ + + + + + + + + + + + + +

z x 1 z 1 x z 1 xz

z xz 1 x yx 1 1 z zx 1 x xz x xy 1 1 x xz xz xyz z

+ + += + = =

+ + + + + +z 1 xz z 1 xz

1.1 x xz xz 1 z 1 x xz

B. Phn tch a thc thnh nhn t

Bi 1

Phn tch tam thc bc hai x2 - 6x + 8 thnh nhn t.

Gii

Cch 1: Tch hng t khng i thnh hai hng t ri a a thc v dng hiu

ca hai bnh phng.

x2 - 6x + 8 =(x - 3)2 - 1 = (x - 3 - 1)(x - 3 + 1) = (x - 4)(x - 2).

Cch 2: Tch hng t bc nht thnh hai hng t ri dng phng php nhm cc

hng t v t nhn t chung.

x2 - 6x + 8 = x2 - 2x - 4x + 8 = x(x - 2) - 4(x - 2) = (x - 2)(x - 4).

Bi 2

Phn tch a thc x3 + 3x2 - 4 thnh nhn t.

Gii

Nhm thy x = 1 l nghim a thc cha nhn t x - 1 ta tch cc hng t

ca a thc lm xut hin nhn t x - 1.

C1: x3+ 3x2- 4 =x3-x2+4x2- 4=x2(x - 1)+4(x2-1)=(x-1)(x2 + 4x + 4)=(x-1)(x+2)2.

C2: x3+3x2- 4 =x3-1+3x2- 3 = (x-1)(x2+x+1)+ 3(x-1)(x+1) = (x-1)(x2+ 4x + 4).

Bi 3



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Phn tch a thc (x+1)(x+3)(x+5)(x+7)+15 thnh nhn t.

Gii

(x +1)(x +3)(x +5)(x +7) +15 = [(x +1)(x +7)][(x +3)(x +5)] +15 = (x2 +8x+7)

(x2+8x +15) +15

t: t = x2+8x+7 x2+8x+15 = t + 8 ta c: t(t + 8) +15 = t2 + 8t +15 =(t +4)2 - 1 = (t + 4 + 1)(t + 4 - 1) = (t + 5)(t + 3).

Vy: (x + 1)(x + 3)(x + 5)(x + 7) + 15 = (x2 + 8x + 12)(x2 + 8x + 10) = (x2 + 6x

+ 2x + 12)(x2 + 8x +10) = (x + 6)(x + 2)(x2 + 8x + 10).

BTVN.

Bi 1

Cho x > y > 0 v 2x2 + 2y2 = 5xy, Tnh:x y

P

x y

+=

. (tng t bi 9)

Bi 2

Cho x + y + z = 0, Chng minh rng: x3 + y3 + z3 = 3xyz. (tng t bi 13)

Bi 3

Cho a + b + c = 0, Chng minh rng: a4 + b4 + c4 =2

1(a2 + b2 + c2 )2. (tng t

bi 14)

Bi 4

Cho a, b, c khc khng v a + b + c = 0.

Chng minh rng: + + =+ + + 2 2 2 2 2 2 2 2 2

1 1 10.

a b c b c a a c b

T: a + b + c = 0 a = - (b + c) a2 = (b + c)2 a2=b2 + c2 + 2bc b2 + c2 -

a2 = - 2bc

Bi 5

Phn tch cc a thc sau thnh nhn t.

a/ 4x2 - 3x - 1

b/ x3 + 6x2 + 11x +6

c/ (x-y)3 + (y-z)3 + (z-x)3

Hng dn: x + y + z = 0 x3 + y3+ z3 = 3xyz



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Chuyn 2:

Bt ng thc - Gi tr ln nht, nh nht

A. Bt ng thc

I. Mt s tnh cht ca bt ng thc

1/ a > b v b > c a > c (t/c bc cu)

2/ a > b a + c > b + c (t/c cng vo hai v cng mt s)

3/ a > b > >

< b v c > d a + c > b + d (t/c cng hai bt ng thc cng chiu)

5/> >

> > >

a b 0ac bd

c d 0(t/c nhn hai bt ng thc dng cng chiu)

6/ a > b > 0 >

>

n n

n n

a b

a b(n nguyn dng)

7/ +> + + +a a

a,b,c Ra b a b c

8/ ++> > > +

a c a a c ca,b,c,d R

b d b b d d

9/ Nu a, b, c l 3 cnh ca tam gic th ta c:

*/ a > 0, b > 0, c > 0.

*/ b - c < a < b + c; a - c < b < a + c; a - b < c < a + b

*/ Nu a > b > c th A > B > C

II. Bi tp

Bi 1

Cho 5 s a, b, c, d, e bt k. CMR: a2 + b2 + c2 + d2 + e2 a( b + c + d +

e)(1).

Gii(1) 4a2 + 4b2 + 4c2 + 4d2 + 4e2 - 4ab - 4ac - 4ad - 4ae 0

(a - 2b)2 + (a - 2c)2 + (a - 2d)2 + (a - 2e)2 0. (pcm)

Bi 2

Cho a + b = 1,Chng minh rng: a/ a2 + b2 1/2, b/ a3 + b3 1/4, c/ a4 +

b4 1/8Gii

a/ T (a - b)2 0 a2 + b2 2ab 2(a2 + b2) a2 + b2 + 2ab = (a + b)2 = 1.



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Vy a2 + b2 1/2.

b/ Ta c a3 + b3 = (a + b)(a2 - ab + b2) = a2 - ab + b2

2(a3 + b3) = 2a2 - 2ab + 2b2 = (a - b)2 + a2 + b2 a2 + b2

m a2 + b2 1/2 2(a3 + b3) 1/2 a3 + b3 1/4. (pcm)

c/ T (a2 - b2)2 0 a4 + b4 2a2b2 2(a4 + b4) a4 + b4 + 2a2b2 = (a2 + b2)2

a4 + b412

(a2 + b2)2 (1).

Mt khc: (a - b)2 0 a2+ b2 2ab 2(a2 + b2) a2 + b2 + 2ab = (a + b)2 = 1

a2 + b2 1/2 (a2 + b2)2 1/4 thay vo (1) ta c a4 + b418

.

Bi 3

Cho a,b > 0, v a + b = 1. Chng minh rng:

a/ + + 1 1

(1 )(1 ) 9a b

; b/ + + +1 1 4

a 1 b 1 3Gii

a/+ + + + +

+ + + 1 1 a 1 b 1 ab a b 1 2

(1 )(1 ) 9 ( )( ) 9 9 1 9a b a b ab ab

1 4ab (a + b)2 4ab ng (pcm).

b/ + + +1 1 4a 1 b 1 33(a + 1 + b +1) 4(a + 1)(b + 1) 9 4(ab + a + b + 1)

9 4ab + 8 1 4ab (a + b)2 4ab ng (pcm)

Bi 4

Cho a, b, c R+. Chng minh rng: < + + + + + > + + + > + + +

a a

a b a b cb b

b c a b cc c

c a a b c

+ + >+ + +a b c

1a b b c c a

.

Mt khc:



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+ <


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( ) + + +

24 4 4 4

4

2 1 2 aba b 0 a b 2 ab ab

a b ab a b.

III/ Bt ng thc Csi(trung bnh cng ln hn hoc bng trung bnh

nhn)

*/ Vi 2 s thc a, b khng m ta c:

+ a b

ab2

, du bng xy ra a = b.

*/ Vi 3 s thc a, b, c khng m ta c:

+ + 3a b c abc3

, du bng xy ra a = b = c.

*/ Vi n s thc a1, a2, ... an khng m ta c:

+ + + 1 2 n n 1 2 na a ... a a a ...an , du bng xy ra a1 = a2 = ... = an .

IV/ Bt ng thc Bunhiacpxki

*/ vi 4 s thc a, b, c, d ta c:

(ab + cd)2 (a2 + c2)(b2 + d2), du bng xy ra =a c

b d.

*/ Vi 6 s thc a, b, c, d, e, f ta c:

(ab + cd + ef)2

(a2

+ c2

+ e2

)(b2

+ d2

+ f2

), du bng xy ra = =a c e

b d f.

*/ vi n cp s thc a1, a2, ... an, b1, b2, ... bn ta c:

(a1b1 +a2b2 + ... + anbn)2 (a12 + a22 + ... + ann)(b12 + b22 + ... + bnn).

Du bng xy ra = = =1 2 n1 2 n

a a a...

b b b.

Bi 8

Cho x, y, z l cc s dng, Chng minh rng:

a/ (x + y)(y + z)(z + x) 8xyz.

b/ + +

1 1 4

x y x y.

c/ + + + +

1 1 1 9

x y z x y z.

Gii



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a/

+ +

+

x y 2 xy

y z 2 yz

z x 2 xz

(x + y)(y + z)(z + x) 8xyz.

b/ + + + +

1 1 4 1 1(x y)( ) 4

x y x y x ym

+ +

x y 2 xy1 1 2

x y xy

+ + 1 1

(x y)( ) 4x y

.

c/ + + + + + + + +

1 1 1 9 1 1 1(x y z)( ) 9

x y z x y z x y z. (lm tng t)

B/ Gi tr ln nht, nh nht

Bi 1

Tm gi tr ln nht ca: P = + +

2

22x 4x 5x 2x 2Gii

Ta c:

P = + + += = + = +

+ + + +

2 2

2 2 2 2

2x 4x 5 2(x 2x 2) 1 1 12 2

x 2x 2 x 2x 2 x 2x 2 (x 1) 1

P ln nht + +2

12

(x 1) 1ln nht, mun vy (x- 1)2 + 1 phi nh nht

m (x- 1)2 + 1 1 (x- 1)2 + 1 nh nht bng 1 x = 1. Khi P = 3

Vy Pmax = 3 x = 1.

Bi 2

Cho x2 + y2 = 1, tm gi tr ln nht v gi tr nh nht ca biu thc: p

= x + yGii

T (x - y)2 0 x2 + y2 2xy 2(x2 + y2) x2 + 2xy + y2 = (x + y)2

Vy 2 (x + y)2 + 2 x y 2

Pmax= 2 x = y =2

2; Pmin= - 2 x = y = -

2

2

Bi 3

Cho x, y > 0 v x + y = 1, Tm gi tr nh nht ca: P = 2 21 1

(1 )(1 )x y

Gii

P = + + + + = = =2 2

2 2 2 2 2 2 2 2

1 1 (x 1)(y 1) (x 1)(x 1)(y 1)(y 1) xy(x 1)(y 1)(1 )(1 )

x y x y x y x y



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=+ + + + + + +

= = = +2 2

xy(x 1)(y 1) (x 1)(y 1) x y xy 1 21

x y xy xy xy. (thay x - 1 = - y, y - 1 = - x)

ta c P nh nht xy

2

nh nht xy ln nht.

M xy = x(1 - x) = - x2 + x = -(x - 1/2)2 + 1/4 1/4 xy ln nht = 1/4 khi x = 1/2 y = 1/2

Vy Pmin =+ =

21 9

1 1.

2 2khi x = y = 1/2.

Bi 4

Tm gi tr ln nht v gi tr nh nht ca: P =+

+

2 2

4

(x 1)

x 1Gii

P =+ + +

= = ++ + +

2 2 4 2 2

4 4 4

(x 1) x 2x 1 2x1

x 1 x 1 x 1

Do (x2 - 1)2 0 x4 + 1 2x2 +

2

4

2x1

x 1 P 2 Pmax= 2 x = 1.

Do 2x2 0, x4 + 1 1 +

2

4

2x0

x 1 P 1 Pmin = 1 =

+

2

4

2x0

x 1 x = 0.

Bi 5

Cho a, b > 0. Tm gi tr nh nht ca; P =+ +(x a)(x b)

x, vi x > 0.

Gii

Ta c:

P =+ + + + +

= = = + + + +2(x a)(x b) x ax bx ab ab

a b x P a b 2 abx x x

.

Vy Pmin = + +a b 2 ab , du bn xy ra = =ab

x x abx .

Bi 6

Tm gi tr nh nht ca: P = + + + +2 21 4x 4x 4x 12x 9Gii

Ta c:

P = ( ) ( )+ + + + = + + = + + 2 22 21 4x 4x 4x 12x 9 1 2x 3 2x 1 2x 3 2x

(1 + 2x) + (3 - 2x) = 4p dng a + b = a + b ab 0. Vy Pmin = 4 (1 + 2x)(3 - 2x) 0



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-1/2 x 3/2.

BTVN

Bi 1

a/ Tm gi tr ln nht ca: P = 5 - 8x - x2.

b/ Tm gi t nh nht ca: P = 4x2 - 4x + 11.

c/ Tm gi tr nh nht ca: P = x - 5 + x- 10.

Hng dn

Ta c: P = x - 5 + x - 10 = x - 5 + 10 - x(x - 5) + (10 - x) = 5

p dng a + b = a + b ab 0. Vy Pmin = 5 (x - 5)(10 - x) 0

5 x 10.

Bi 2

Cho x, y R, Chng minh rng: x2 + y2 + 1 xy + x + y.

Bi 3

Cho a, b, c, d R+.

Chng minh rng :+ + + +< + + + 0).

*/ +

= = +

m m( a b) m m( a b),

a b a ba b a b

II/ Bi tp

Bi 1

Tnh gi tr cc biu thc sau:a/ A = 6 48 2 27 4 75 b/ B =

+ 148 2 75 108 1477

Gii

a/ Ta c: A = = = = 6 48 2 27 4 75 6 16.3 2 9.3 4 25.3 24 3 6 3 20 3 2 3

b/ Ta c: B = + = + = 1 1

48 2 75 108 147 4 3 2.5 3 6 3 .7 3 3

7 7Bi 2

Trc cn thc mu:

a/ A = + +1 1

5 2 5 2b/ B =

+ + +

4

3 5 2 2 5c/ C =

+ +3 32

2 2 2 4

Gii



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a/ A =+

+ = + = +1 1 5 2 5 2 2 5

3 3 35 2 5 2

b/ B =+ + + +

= = =+ + ++ + +

2

4 4(3 5 2 2 5) 3 5 2 2 5

(3 5) (2 2 5) 3 53 5 2 2 5

+ + += =2(3 5)(3 5 2 2 5) 4 (3 5) (2 2 5)

4 4

c/ t =3 2 a C =

= = = = = + + + + + +

3 23 3

4 3 2 2 33 3

2 a a a(a 1) a a4 2

a a a a a 1 a 1 2 12 2 2 4

Bi 3

Rt gn biu thc cha cn:

a/ A = + 15 6 6 33 12 6b/ B = +8 2 15 8 2 15

c/ C = + 4 7 4 7

d/ D = + + + +4 10 2 5 4 10 2 5

e/ E = + + 4 449 20 6 49 20 6

f/ F = + + + ++ + + +

1 1 1 1...

1 5 5 9 9 13 2001 2005Gii

a/ A = + = + + + =15 6 6 33 12 6 9 6 6 6 9 12 6 24

= + + = + + =2 2(3 6) (3 2 6) 3 6 2 6 3 3 6.

b/ B = + = + + + =8 2 15 8 2 15 5 2 15 3 5 2 15 3

+ = + = 2 2( 5 3) ( 5 3) 5 3 ( 5 3) 2 3.

c/ C = + + + ++ = = 8 2 7 8 2 7 7 2 7 1 7 2 7 1

4 7 4 72 2 2 2

+ + = = =

2 2( 7 1) ( 7 1) 7 1 7 12.

2 2 2 2

d/ Do D > 0 nn D = 2D

D2 = + + + + = + + + +

2

4 10 2 5 4 10 2 5 8 2 (4 10 2 5)(4 10 2 5)



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= + = + + = + = + = +28 2 6 2 5 8 2 5 2 5 1 8 2 ( 5 1) 8 2 5 2 6 2 5

Vy: D = + = + = +26 2 5 ( 5 1) 5 1

e/ Ta c: + = + + = + = + = +2 2 2 449 20 6 25 20 6 24 (5 2 6) [( 3 2) ] ( 3 2)

= + = = = 2 2 2 449 20 6 25 20 6 24 (5 2 6) [( 3 2) ] ( 3 2)

Vy E = + + =3 2 3 2 2 3.

f/ F = + + + + =5 1 9 5 13 9 2005 2001 2005 1...4 4 4 4 4

.

Bi 4

Rt gn cc biu thc sau:

a/ A =+ +

x 4 x 4 x 4 x 4

b/ B = + 2 2 2 2x 2 x 1 x 2 x 1

c/ C = + + 2 22x 1 2 x x 2x 1 2 x xGii

a/ A = + + = + + + +x 4 x 4 x 4 x 4 x 4 4 x 4 4 x 4 4 x 4 4

= + + = + + 2 2( x 4 2) ( x 4 2) x 4 2 x 4 2

Nu x 4 2 x 4 4 x 8 th A = +x 4 2+ = x 4 2 2. x 4.

Nu < < < < <


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Bi 5

Tm iu kin cc biu thc sau c ngha v rt gn:

a/ + +

2x 2 x 1 x 2 x 1 1

(1 )x 1x 4(x 1)

b/ +

31 1 x x

x x 1 x x 1 1 x

c/+

+ +21 x 1

:x x x x x x

d/+ + + +

2 x x 1 x 2( ):

x x 1 x 1 x x 1

e/ + + + + + x 2 x 1 x 1( ): 2x x 1 x x 1 1 xGii

a/ K:> > >

+ > > 2 2

x 1 x 1 x 1

x 2x 4x 4 0 (x 2) 0.

A = + + + +

=

2 2

2 2

x 2 x 1 x 2 x 1 1 ( x 1 1) ( x 1 1) x 2(1 ) .

x 1 x 1x 4(x 1) (x 2)

= 22 x 2. x 1(x 2) .

Nu x > 2 A = =2

x 1

Nu 1< x < 2 A = =2

1 x

Vy: A =

> <

x 1x 1

x 1 0.

B = +

= +

31 1 x x x x 1 x x 1 x( x 1)

1 1x x 1 x x 1 1 x 1 x

= + = 2 x 1 x x 2 x 1 .



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c/

> + + +

2

x 0

x x 0 x 0

x 1x x x x 0

x 1 0

.

t = = 2x a x a

C =+ + + + += =

+ + + +

3 2 2

4 32

1 x 1 1 a a a a(a a 1):

a a a 1 a(a 1)(a 1)x x x x x x

=+ +

= =+ + +

2

2 2

a a 1 1 1

(a 1)(a a 1)(a 1) a 1 x 1.

d/ K:

x 0x 0

x 1 0

x 1x x 1 0

.

t = = 2x a x a

D =+ + + + + =

+ + +

2 2

3

2 x x 1 x 2 2a a 1 a a 1( ): ( )( )

a 1 a 1 a 2x x 1 x 1 x x 1

=+ + + + + +

= = = + + + + + +

2 2

2

a(a 2) (a a 1) a a 1 a 1 1 1.

(a 1)(a a 1) a 2 (a 1)(a 2) a 2 x 2.

e/ K:

x 0 x 0x x 1 0

x 11 x 0

t = = 2x a x a

E =+ +

+ + = + + + + +

2

3 2

x 2 x 1 x 1 a 2 a 1 2( ): ( )

2 a 1 a a 1 a 1 a 1x x 1 x x 1 1 x

=+ + + + +

= = = + + + + + + + +

2 2 2

2 2 2

a 2 a(a 1) (a a 1) 2 a 2a 1 2 2 2

(a 1)(a a 1) a 1 (a 1)(a a 1)a 1 a a 1 x x 1

.

Bi 6

Chng minh rng cc biu thc sau l mt s nguyn.

a/ A = + + +4 5 3 5 48 10 7 4 3

b/ B = + + + ( 3 1) 6 2 2 3 2 12 18 128



19/63


c/ C = + +

2 3 5 13 48

6 2Gii

a/ Ta c: + = + + = + = 27 4 3 (2 3) 10 7 4 3 10(2 3) 20 10 3

+ = = =

+ = =

248 10 7 4 3 48 20 10 3 28 10 3 (5 3)

5 48 10 7 4 3 5(5 3) 25 5 3

Vy A = + =4 5 3.

b/ Ta c: = = 218 128 18 8 2 (4 2)

+ + = + + = + = + 22 12 18 128 2 12 4 2 4 2 3 ( 3 1)

+ + = + = + = + = +6 2 2 3 ( 3 1) 6 2 4 2 3 6 2( 3 1) 4 2 3 3 1

Vy: B = + = =( 3 1)( 3 1) 3 1 2.

c/ Ta c: + = + = + + = + + = + 213 48 13 4 3 12 4 3 1 (2 3 1) 13 4 3 2 3 1

+ = = = + = 25 13 48 5 2 3 1 4 2 3 ( 3 1) 5 13 48 3 1

+ + = + + = + + + =3 5 13 48 3 3 1 2 3 2 3 5 13 48

+ = + = + =22 2 3 8 4 3 ( 6 2) C 1.

BTVN

Bi 1

Rt gn biu thc cha cn.

a/ A = + 4 15 4 15 2 3 5 b/ B = 5 3 29 12 5

c/ C = + (5 2 6)(49 20 6) 5 2 6

9 3 11 2 d/ D = + + ++ + +1 1 1

...2 3 3 4 1998 1999

Bi 2

Trc cn thc mu.

a/ A = +3 36

2 2 2 4b/ B =

+ +3 3

2

4 2 2



20/63


Chuyn 4

Phng trnh bc nht - th hm s bc nht - H phng trnh bc

nht

I/ Phng trnh bc nht

N: L phng trnh c dng: ax + b = 0, trong a, b l cc s thc, x l n.Cch gii:

Phng trnh ax = -b.

Nu a 0 x = -b/a

Nu a = 0 0x = -b

Nu b = 0 PT v s nghim

Nu b 0 PT v nghim

II/ Bi tp

Bi 1

Gii v bin lun cc phng trnh sau:

a/ mx + 2(x - m) = (m + 1)2 + 3 (1) b/ 3(m + 1)x + 4

= 2x + 5(m + 1) (2)

c/ m2(x + 1) = x + m (3) d/ + =

x m x 32

x 2 x

(4)Gii

a/ (1) (m + 2)x = m2 + 4m + 4 (m + 2)x = (m + 2)2

Nu m + 2 0 m -2 phng trnh c nghim: x = m + 2.

Nu m + 2 = 0 m = -2 0x = 0 0 phng trnh c v s nghim x R.

b/ (2) (3m + 1)x = 5m + 1

Nu 3m + 1 0 m -1/3 phng trnh c nghim:+

= +5m 1

x 3m 1

Nu 3m + 1 = 0 m = -1/3 phng trnh c dng: 0x = -2/3 PTVN.

c/ (3) (m2 - 1)x = m - m2 (m2 - 1)x = m(1 - m).

Nu m2 - 1 0 phng trnh c nghim: = +m

xm 1

Nu m2 - 1 = 0 m = 1.

Nu m = 1 PT c dng: 0x = 0 PT c VSN

Nu m = -1 PT c dng: 0x = -2 PTVN



21/63


d/ K: x 0 v x 2.

(4) x(x - m) + (x - 2)(x - 3) = 2x(x - 2) (m + 1)x = 6

Nu m + 1 = 0 m = -1 (4) c dng: 0x = 6 PTVN

Nu m + 1 0 m -1 (4) = +6

x 0m 1 (Do K m 2 +6

2 m 2m 1 )

Kt lun: Nu m -1 m 2 phng trnh c nghim: =+6

xm 1

Nu m = -1 m = 2 phng trnh v nghim.

Bi 2

Cho phng trnh: (m + 1)2x + 1 - m = (7m - 5)x. (1)

a/ Tm m phng trnh v nghim b/ Tm m

phng trnh c nghimGii

(1) ( m2 - 5m + 6)x = m - 1 (m - 2)(m + 3)x = m - 1.

a/ Phng trnh v nghim + =

= =

(m 2)(m 3) 0m 2 m 3.

m 1 0

b/ phng trnh c nghim (m - 2)(m + 3) 0 m 2 m -3.

III/ H phng trnh bc nht

Bi 3

Cho h phng rnh:+ =

+ =

2x my 1 (1)

mx 2y 1 (2).

a/ Gii h khi m = 1

b/ Gii v bin lun h phng trnh

c/ Tm cc s nguyn m h c nghim duy nht (x ; y) vi x, y l cc

s nguyn

d/ Tm cc s nguyn m h c nghim (x ; y) vi x, y l cc s nguyn

dngGii

a/ khi m = 1 ta c h + = + = = =

+ = + = + = =

2x y 1 4x 2y 2 3x 1 x 1 3

x 2y 1 x 2y 1 x 2y 1 y 1 3

b/ T (1) v (2) 2x + my = mx + 2y (m - 2)(x - y) = 0.

Nu m = 2 h v s nghim

Nu m 2 x = y thay vo phng trnh (1) ta c: (m + 2)x = 1.



22/63


Nu m = -2 h v nghim

Nu m -2 h c nghim duy nht: x = y = 1/(m + 2)

c/ khi m 2 v m -2 th h c nghim duy nht: x = y = 1/(m + 2). Nghim

ny l s nguyn 1/(m + 2) l s nguyn + = = + = = m 2 1 m 1m 2 1 m 3

.

d/ / khi m 2 v m -2 th h c nghim duy nht: x = y = 1/(m + 2). Nghim

ny l s nguyn dng 1/(m + 2) l s nguyn dng m + 2 l c s nguyn

dng ca 1 m + 2 = 1 m = -1.

Bi 4

Cho h phng rnh:

= = +

(m 1)x my 3m 1 (1)

2x y m 5 (2)

a/ Tm m h c nghim duy nht (x ; y) m S = x 2 + y2 t gi tr

nh nht.

b/ Tm m h c nghim duy nht (x ; y) m P = xy t gi tr ln

nht.Gii

T (2) y = 2x - m - 5 thay vo (1) (m - 1)x - 2mx + m2 + 5m = 3m -1

(m + 1)x = m2 + 2m + 1 (m + 1)x = (m + 1)2.

H c nghim duy nht m -1, khi : x = m + 1, y = m - 3.

a/ S = x2 + y2 = (m+1)2 + (m-3)2 = 2m2 - 4m + 10 = 2(m - 1)2 + 8. Smin = 8

m = 1.

b/ P = xy = (m + 1)(m - 3) = m2 -2m -3 = (m - 1)2 - 4. Pmin = -4 m = 1.

Bi 5

Gii h phng trnh:

+ + = + + =

x y 2x y7 (1)7 17

4x y y 715 (2)

5 19Gii

(1) 17(x - y) + 7(2x + y) = 7.7.17 31x - 10y =833.

(2) 19(4x + y) + 5(y - 7) = 19.5.15 19x + 6y = 365.

Vy h phng trnh = = =

+ = + = =

31x 10y 833 93x 30y 2499 x 23

19x 6y 365 95x 30y 1825 y 12

.

Bi 6



23/63


Gii h phng trnh:

+ + = + + = + + =

x y z 1 (1)

x 2y 4z 8 (2)

x 3y 9z 27 (3)

Gii

H:+ + = + + = + + = = + + = + = + = =

+ + = + = = =

x y z 1 x y z 1 x y z 1 x 6x 2y 4z 8 y 3z 7 y 3z 7 y 11

x 3y 9z 27 y 5z 19 2z 12 z 6

IV/ th hm s bc nht

th hm s y = ax + b (a 0) l ng thng i qua hai im A(0;b) v

B(-b/a; 0).

Bi 7

V th cc hm s sau:a/ y = 2x - 1 b/ y = x 1 c/ y = +22 x 2x 1 d/ y = + + x 1 x 2

e/ + =x y 1BTVN

Bi 1 Gii v bin lun cc phng trnh sau:

a/ m2x = 9x + m2 - 4m + 3 b/

+ + =+

x m x 22x 1 x

Bi 2 Cho h phng trnh:+ =

=

x my 2

mx 2y 1.

a/ Gii h khi m = 2

b/ Tm s nguyn m h c nghim duy nht (x ; y) vi x, y l cc s nguyn

c/ Tm s nguyn m h c nghim duy nht (x ; y) m x > 0 v y < 0

Bi 3 V th cc hm s: a/ y = 2x - x + 3 b/ y = x - 1 - x + 2

Bi 4 Gii h phng trnh:

+ + = + + = + + =

x 2y 3z 11

2x 3y z 2

3x y 2z 3

Hng dn Cng 3 phng trnh ta c: x + y + z = 2. x = -2, y = -1, z = 5.



24/63


Bi 5 Gii h phng trnh:

+ = + = =

+

3z 2 (1)

2x y

2y 3z 4 (2)

2 3y (3)

2x y 2

Hng dn t t = +1

2x ythay vo (1) v (3) ta c:

+ = =

3t z 2

32t y

2

2z + 3y = -1/2 (4).

T (2) v (4) ta c: x = 1/4, y = 1/2, z = -1.



25/63


Chuyn 5

Phng trnh bc 2, nh l vit - Phng trnh bc cao

I/ Phng trnh bc 2

N: Phng trnh bc 2 l phng rnh c dng: ax2 + bx + c = 0. (a 0)

Trong : a, b, c l cc s thc, x l n.

Cch gii:

Tnh bit thc = b2 - 4ac

Nu < 0 phng trnh v nghim.

Nu = 0 phng trnh c nghim kp: x = -b/2a.

Nu > 0 phng trnh c 2 nghim phn bit: + = =1 2b b

x ; x4a 4a

Ch : Nu b = 2b' th c th tnh ' = b'2 - ac

Nu ' < 0 phng trnh v nghim.

Nu ' = 0 phng trnh c nghim kp: x = -b '/a.

Nu ' > 0 phng trnh c 2 nghim phn bit: + = =' ' ' '

1 2

b bx ; x

2a 2a

II/ nh l Vit

Nu phng trnh bc 2 c hai nghim phn bit hoc khng th ta c:S = x1 + x2 = -b/a; P = x1x2 = c/a.

Ch :

Nu phng trnh bc 2 c a + b + c = 0 th x1 = 1; x2 = c/a.

Nu phng trnh bc 2 c a - b + c = 0 th x1 =-1; x2 = -c/a.

III/ Bi tp

Bi 1

Cho phng trnh: x2

- 4x + m + 1 = 0.a/ Gii phng trnh khi m = 2

b/ Tm m phng trnh c nghim

c/ Tm m phng trnh c 2 nghim x1, x2 tho mn: x12 + x22 = 10

d/ Tm m phng trnh c 2 nghim x1, x2 tho mn: x13 + x23 = 34Gii

a/ Khi m = 2 PT x2 - 4x + 3 = 0 do a + b + c = 0 x1 = 1, x2 = 3.

b/ ' = 4 - m - 1 = 3 - m, phng trnh c nghim 3 - m 0 m 3.

c/ phng trnh c 2 nghim th phi c 0 m 3.



26/63


Khi : x12 + x22 = 10 (x1 + x2)2 - 2x1x2 = 10 16 - 2(m + 1) = 10 m = 2

d/ phng trnh c 2 nghim th phi c 0 m 3.

x13 + x23 = 34 (x1 + x2)[(x1 + x2)2 -3x1x2] =34 4[16 -3(m + 1)] =34 m +1

=10 m = 9

Bi 2

Cho phng trnh: x2 - 2(m - 1)x - 3 - m = 0.

a/ Chng minh rng phng trnh lun c nghim vi mi m.

b/ Tm phng trnh c mt nghim x = 2, tm nghim kia

c/ Tm m phng trnh c 2 nghim x1 , x2 tho mn x12 + x22 10

d/ Tm m phng trnh c nghim x1 , x2 sao cho P = x12 + x22 t gi

tr nh nhtGii

a/ ' = m2 - 2m + 1 + m + 3 = m2 - m + 4 = (m- 1/2)2 + 15/4 > 0 vi mi m

th phng trnh lun c nghim.

b/ x = 2 thay vo phng trnh ta c: 5m = 5 m = 1. Khi phng trnh c

dng: x2 - 4 = 0 x = 2 x = -2.

c/ x12 + x22 10 (x1 + x2)2 - 2x1x2 10 [2(m - 1)]2 + 2(m + 3) 10

4m2

-8m + 4 + 2m + 6 10 4m2

- 6m 0 m(2m - 3) 0 m 3/2 m 0.

d/ P = x12 + x22 = (x1 + x2)2 - 2x1x2 = [2(m - 1)]2 + 2(m + 3) = 4m2 - 6m + 10 =

(2m - 3/2)2 + 31/4 Pmin = 31/4 m = 3/4.

Bi 3

Cho phng trnh: x2 - 2mx + 2m -1 = 0.

a/ Chng minh rng phng trnh lun c nghim vi mi m.

b/ Tm m phng trnh c 2 nghim x1 , x2 tho mn 2x12

+ 2x22

-5x1x2 = 27.

c/ Tm m sao cho phng trnh c nghim ny bng hai nghim kia.

d/ Tm m phng trnh c 2 nghim x1 , x2 tho mn: x1 = x22Gii

a/ ' = m2 - 2m + 1 = (m + 1)2 0 vi mi m phng trnh lun c nghim.

b/ 2x12 + 2x22 - 5x1x2 = 27 2[(x1 + x2)2 - 2x1x2] - 5x1x2 = 27 2(x1 + x2)2 - 9x1x2

= 27 8m2

- 9(2m + 1) = 27 8m2

- 18m - 18 = 0 4m2

- 9m - 9 = 0 m = 3 m = -3/4.



27/63


c/ Gi s phng trnh c 2 nghim: x1 = 2x2 ta c:

x1 + x2 = 3x2 =2m x2 =2m/3 (1) v x1x2 = 2x22 = 2m - 1x22 = (2m - 1)/2 (2).

T (1) v (2) 4m2/9 = (2m - 1)/2 8m2 - 18m + 9 = 0 m = 3/4 m = 3/2

d/ Ta c: x = m + m + 1 = 2m + 1 x = m - m - 1 = -1

Nu x1 = 2m + 1, x2 = -1 th ta c: 2m + 1 = 1 m = 0

Nu x1 = -1, x2 = 2m + 1 th ta c: -1 = (2m + 1)2 v l. Vy m = 0.

Bi 4

Cho phng trnh: (m - 1)x2 + 2(m - 1)x - m = 0.

a/ Tm m phng trnh c nghim kp, tm nghim kp ny

b/ Tm m phng trnh c 2 nghim phn bit tri du

c/ Tm m phng trnh c 2 nghim phn bit u m

d/ Tm m phng trnh c 2 nghim phn bit u dngGii

a/ Phng rnh c nghim kp m 1 v ' = 0 m2 - 2m + 1 + m2 - m = 0

2m2 - 3m + 1 = 0 (m - 1)(2m - 1) = 0 m = 1 m = 1/2

Vy m = 1/2 th phng trnh c nghim kp: x = 1.

b/ Phng trnh c 2 nghim phn bit tri du

> < > >

'

1 2

m 1m 1 m 1m 1/2 m 0

0 (m 1)(2m 1) 0m 1m 0

x x 0 m0 m 1

m 1

.

c/ Phng trnh c 2 nghim phn bit u m

> > >

< <


28/63


> > < < >> > + > >

'

1 2

1 2

m 1 m 1m 1 (m 1)(2m 1) 0 m 1/2

0 m 0 m 10x x 0 m 1 2 0

2(m 1)x x 0

0m 1

Loi

Vy khng tn ti m phng trnh c 2 nghim phn bit u dng.

Bi 5

Cho phng trnh: x2 - (2m - 3)x + m2 - 3m = 0.

a/ Chng minh rng phng trnh lun c 2 nghim khi m thay i.

b/ Tm m phng trnh c 2 nghim x1, x2 tho mn: 1 < x1 < x2 < 6.Gii

a/ = 4m2 - 12m + 9 - 4m2 + 12m = 9 > 0 phng trnh lun c 2 nghim.

b/ x1 = = 2m 3 3 m 32

; x2 = + =2m 3 3 m2

Vi mi m ta lun c: m - 3 < m 1 < m - 3 < m < 6 4 < m < 6.

Bi 6

Cho phng trnh: 3x2 - mx + 2 = 0. Tm m pt c 2 nghim tho mn:

3x1x2 = 2x2 - 2.Gii

K:

= = = =

= = = + = + = =

2

1 2 2 2 2

1 2 1 2 1

1 2 1 2

m 24 0 m 2 6 m 2 6 m 2 6 m 2 6

3x x 2x 2 2 2x 2 x 2

x x 2/3 x x 2/3 x 1/3

x x m/3 x x m/3 m 7

Bi 7

Gi a, b l nghim ca phng trnh: x2 + px + 1 = 0

c, d l nghim ca phgn trnh: x2 + qx + 1 = 0a/ Chng minh rng: (a - c)(a - d)(b - c)(b - d) = (p - q)2

b/ Chng minh rng: (a - c)(b - c)(a + d)(b + d) = q2 - p2

Gii

Theo nh l Vit ta c:+ = + =

= =

a b p c d q

ab 1 cd 1.

a/ VT = (a - c)(a - d)(b - c)(b - d) = (a2 - ad - ac + cd)(b2 - bc - bd + cd) =

[a2 - a(c + d) + cd][b2 - b(c + d) + cd] = (a2 + aq + 1)(b2 + bq + 1) =

a2b2 + a2bq + a2 +ab2q + abq2 + aq + b2 + bq + 1 =



29/63


1 + aq + bq + q(a + b) + [(a + b) 2 - 2ab] + q2 + 1 =

2 + q(a + b) - pq + p2 - 2 + q2 + 1 = p2 - 2pq + q2 = (p - q)2 = VP.

b/ VT = (a - c)(b - c)(a + d)(b + d) = [ab - c(a + b) + c 2][ab + d(a + b) + d2] = (1

+ cp + c2)(1- dp + d2) = 1- dp + d2 + cp - cdp2 + cd2p + c2 - c2dp + c2d2 =

= 1- dp + d2 + cp - p2 + dp + c2 - cp + 1 = (c + d)2 - 2cd - p2 + 2 = q2 - p2 = VP.IV/ Phng trnh bc cao

Bi 8

Gii cc phng trnh sau:

a/ x3 - 2x2 - x + 2 = 0

b/ x4 - 3x3 + 6x2 + 3x + 1 = 0

c/ x4 + 2x3 - 6x2 + 2x + 1 = 0

d/ (x2 - 3x + 1)(x2 - 3x + 2) = 2e/ (x + 9)(x + 10) (x + 11) - 8x = 0

f/ (x + 2)2 + (x + 3)3 + (x + 4)4 = 2Gii

a/ Nhm thy x = 2 l nghim phn tch VT lm xut hin x - 2

x3- 2x2 - x + 2 = 0 x2(x- 2)- (x- 2) = 0 (x- 2)(x2 - 1) = 0 x = 0 x = 1.

Cch khc:

x3 - 2x2 - x + 2 = 0 x3 - 8 - (2x2 - 8) - (x - 2) = 0 (x - 2)(x2 + 2x + 4 - 2x - 4 -1) (x - 2)(x2- 1) = 0 x = 0 x = 1.

b/ Do x = 0 khng phi l nghim chia c hai v cho x2 0 ta c:

+ + = + = 2 2

2 2

3 1 1 1x 3x 6 0 x 3 x 6 0

x x x x

t: = + = +2 221 1

x t x t 2x x

, thay vo phng trnh ta c:

t2 - 3t - 4 = 0 (t + 1)(t - 4) = 0 t = -1 t = 4.

Vi : t = -1

= = + = +=

12

2

1 5x

1 2x 1 x x 1 0x 1 5

x2

.

Vi : t = 4 =

= = = +

12

2

x 2 51x 4 x 4x 1 0

x x 2 5.

c/ Do x = 0 khng phi l nghim chia c hai v cho x2 0 ta c:



30/63


+ + = + = 2 2

2 2

2 1 1 1x 2x 6 0 x 2 x 6 0

x x x x

t: = + = +2 221 1

x t x t 2x x


t2 - 2t - 4 = 0 = + = t 1 5 t 1 5 (khng tm c x)Cch khc:

x4 + 2x3 - 6x2 + 2x + 1 = 0 (x4 - 2x2 + 1) + (2x3- 4x2 + 2x) = 0

(x2 - 1)2 + 2x(x - 1)2 = 0 (x - 1)2[(x + 1)2 + 2x] = 0 =

+ + =2

x 1

x 4x 1 0

e/ t t = x2 - 3x + 1 phng trnh c dng: t(t + 1) = 2 t2 + t - 2 = 0 t =

1 t = -2.

Vi: t = 1 x2 - 3x = 0 x = 0 x = 3.

Vi: t = -2 x2 - 3x + 3 = 0, VN

f/ t: x+ 10 = t (t - 1) t (t + 1) - 8(t - 10) = 0 t3 - 9t + 80 = 0

(t + 5)(t2 - 5t + 16) = 0 t = -5 x = -15.

g/ t: x + 4 = t (t - 2)2 + (t - 1)3 + t4 = 2

(t2 - 4t + 4) + (t3 - 3t2 + 3t - 1) + t4 = 2 (t2 -1)(t2 + t - 1 = 0

=+

=

01

01

2

2

tt

t

.

Bi 9

Cho phng trnh: x3 - 2x2 + (m + 1)x - m = 0.

a/ Chng minh rng: phng trnh lun c nghim x = 1 vi mi m

b/ Tm m phng trnh c ng 2 nghim

c/ Gii v bin lun phng trnh theo mGii

a/ Thay x = 1 vo phng trnh ta thy lun ng x = 1 l nghim vi mi mb/ Pt (x - 1)( x2 - x + m) = 0

t: f(x) = x2 - x + m PT c ng 2 nghim



31/63


TH1: f(x) = 0 c nghim kp x 1

4

1

0

041

0)1(

0=

=

=m

m

m

f.

TH2: f(x) = 0 c 2 nghim phn bit trong mt nghim phi bng 1

> > 1/4 PT c mt nghim x = 1.

Nu = =

=

0 1 4m 0 1m

f(1) 0 m 0 4 PT c hai nghim x = 1 x = 1/2.

Nu > > >


32/63


t: = + = +2 221 1

x t x t 2x x


2t2 - t - 1 = 0 (t - 1)(2t +1) = 0 t = -1 t = -1/2.

Vi : t = 1

= = = +=

12

2

1 5x

1 2x 1 x x 1 0x 1 5

x2

.

Vi : t = -1/2

= = + = +=

12

2

1 17x

1 1 4x 2x x 2 0x 2 1 17

x4

.



33/63


Chuyn 6

Gii phng trnh, bt phng trnh cha gtt v cn thc

I/ Gii phng trnh cha GTT

*/ Dng c bn.

A= B = = 2 2

B 0 B 0A BA B

A= B A2 = B2 A = B

*/ Dng khng c bn

- Dng nh ngha: A=


34/63


c/ t: t = x2 - 1 x2 - 4 = t - 3

x2 - 1+ x2 - 4 = 3 t + t - 3 =3 (*)

Nu: t 3 (*) 2t = 6 t = 3 x2 = 4 x = 2

Nu: 0 t < 3 (*) 3 = 3 PT c v s nghim 0 t < 3 0 x2 -1 < 3

1 x2 < 4


35/63


f/++ + = x 3x 2 x 1 x 2 x 12

g/ + + + =x 2 3 2x 5 x 2 2x 5 2 2Gii

a/ + = ==

+ = = =2 2

x 1x 1 0 x 1x 1 x 1 x 3x 0

x 1 (x 1) x 3x 0x 3

b/ K:

+

1 x 0 x 14 x 1

4 x 0 x 4

+ + = + + = + = 1 x 4 x 3 5 2 (1 x)(4 x) 9 (1 x)(4 x) 2

+ = = = 2x 3x 0 x 0 x 3

c/ K:

+

1 x 0 x 12 x 1

2 x 0 x 2

+ = = + + = + + + 1 x 2 x 1 1 x 1 2 x 1 x 3 x 2 2 x

+ = +2 x (x 1)

+ = =

+ = + + + = + =

2 2

x 1

1 5x 1 0 x 1 1 5x x222 x x 2x 1 x x 1 0 1 5

x2

d/ + + + + = + + = + + =2 2 2 2x 2x 1 x 4x 4 3 (x 1) (x 2) 3 x 1 x 2 3

+ + = + + = + +1 x x 2 3 1 x x 2 (1 x) (x 2)

p dng: + = + a b a b a.b 0 ta c: (1 - x)(x + 2) 0 -2 x 1

e/ + + + + =x 3 4 x 1 x 8 6 x 1 5

K: x 1

+ + + + = + + + + =x 3 4 x 1 x 8 6 x 1 5 x 1 4 x 1 4 x 1 6 x 1 9 5

+ + = + + = 2 2( x 1 2) ( x 1 3) 5 x 1 2 x 1 3 5

+ + = + + = + + x 1 2 3 x 1 5 x 1 2 3 x 1 ( x 1 2) (3 x 1)

p dng: + = + a b a b a.b 0 ta c:

+ ( x 1 2)(3 x 1) 0 2 x 1 3 5 x 10.



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f/+

+ + =x 3

x 2 x 1 x 2 x 12

K: x 1

+ ++ + = + + + + =

x 3 x 3x 2 x 1 x 2 x 1 x 1 2 x 1 1 x 1 2 x 1 1

2 2

+ + + + = + + =2 2 x 3 x 3( x 1 1) ( x 1 1) x 1 1 x 1 12 2

(*)

Nu:+ = = + + 2x 3x 1 1 x 2 (*) 2 x 1 16x 16 x 6x 92

+ = = =2 2x 10x 25 0 (x 5) 0 x 5.

Nu:+ < < = =x 3x 1 1 x 2 (*) 2 x 1.2

g/ + + + =x 2 3 2x 5 x 2 2x 5 2 2

K: x 5/2.

+ + + = x 2 3 2x 5 x 2 2x 5 2 2

+ + + =2x 4 6 2x 5 2x 4 2 2x 5 4

+ + + + = 2x 5 6 2x 5 9 2x 5 2 2x 5 1 4

+ + =2 2( 2x 5 3) ( 2x 5 1) 4

+ + = + + = 2x 5 3 2x 5 1 4 2x 5 3 1 2x 5 4

+ + = + + 2x 5 3 1 2x 5 ( 2x 5 3) (1 2x 5)

p dng: + = + a b a b a.b 0 ta c:

+ ( 2x 5 3)(1 2x 5) 0 1 2x 5 0 2x 5 1 x 3.

Vy: 5/2 x 3

Bi 3

Gii cc phng trnh sau:

a/ + = + + 2 23x 2x 2 x x 1 x

b/ + + =x x(5 2 6) (5 2 6) 10Gii

a/ + = + + 2 23x 2x 2 x x 1 x

K: x2 + x 0 x 0 x -1



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PT + = + + = +2 2 2 23x 3x 1 2 x x 3(x x) 1 2 x x

t: + = 2x x t (t 0) ta c: 3t2 - 2t - 1 = 0 t = 1 t = -1/3 (loi)

t = 1 + = + = =2 2 1 5x x 1 x x 1 0 x

2

b/ Do: + =(5 2 6)(5 2 6) 1 t: = >x(5 2 6) t (t 0) PT + =1

t 10t

= + =

= +

2 t 5 2 6t 10t 1 0t 5 2 6

Vi = = =xt 5 2 6 (5 2 6) 5 2 6 x 2

Vi = + = + = = x x 1t 5 2 6 (5 2 6) 5 2 6 (5 2 6) (5 2 6) x 2

Vy phng trnh c hai nghim: x = 2.

Bi 4

Gii phng trnh:

a/ + = + + 2 23x 2x 2 x x 1 x

b/ + + = +2x 4x 5 2 2x 3

c/ + + = + + + 2x 10x 21 3 x 3 2 x 7 6

d/ + + + =2 23x 12x 16 y 4y 13 5

e/ + + + + + = 2 2 23x 6x 7 5x 10x 14 4 2x x

f/ ( ) ( ) + = + +2 2 27

x 3x x 2x 2 x 4x 52

g/ + = +22x 3 5 2x 3x 12x 14Gii

a/ K: x2 + x 0 x 0 x -1

+ = + + + = + +2 2 2 23x 2x 2 x x 1 x 3(x x) 2 x x 1

t + = 2x x t(t 0) PT 3t2 - 2t - 1 = 0 t = 1 t = -1/3 (loi)

= + = + = =2 2 1 5t 1 x x 1 x x 1 0 x2

b/ K: 2x + 3 0 x -3/2



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+ + = + + + + + + + 2 2x 4x 5 2 2x 3 (x 2x 1) (2x 3) 2 2x 3 1

+ = + + + = = + =

22 2

2

(x 1) 0(x 1) ( 2x 3 1) 0 x 1

( 2x 3 1) 0.

c/ K:+ +

x 3 0x 3

x 7 0

+ + = + + + + + + + + =2x 10x 21 3 x 3 2 x 7 6 (x 3)(x 7) 3 x 3 2 x 7 6 0

( ) ( ) ( ) ( ) + + + = + + = x 3 x 7 3 2 x 7 3 0 x 7 3 x 3 2 0

+ = + = = + =+ =

x 7 3 0 x 7 9x 2

x 3 4x 3 2 0.

d/ Do: 3x2 -12x + 16 = 3(x - 2)2 + 4 4 + 23x 12x 16 2

y2 - 4y + 13 =(y - 2)2 + 9 9

+ + + 2 23x 12x 16 y 4y 13 5 PT = = = =

2

2

3(x 2) 0 x 2

y 2(y 2) 0

e/ + + + + + = 2 2 23x 6x 7 5x 10x 14 4 2x x

( ) ( ) ( ) + + + + + = +2 2 2

3 x 1 4 5 x 1 9 5 x 1

M: ( ) ( )+ + + + + + =2 2

3 x 1 4 5 x 1 9 4 9 5 cn 5 - (x+1)2 5

nn ta c: (x+1)2 = 0 x = -1.

f/ Ta c:

x2 - 2x + 2 = (x- 1)2 + 1 > 0

x2 - 4x + 5 = (x- 2)2 + 1 > 0

( ) ( ) ( ) ( ) + + +

+ = + +

2 2

2 2 2

x 2x 2 x 4x 57x 3x x 2x 2 x 4x 52 2

PT x2- 2x + 2 = x2 - 4x + 5 x = 3/2.

g/ K: 3/2 x 5/2

VP = 3 x2 - 12x + 14 = 3(x - 2)2 + 2 2, du bng xy ra x = 2.

VT2 = + + + + 2 2 2 2( 2x 3 5 2x) (2x 3 5 2x)(1 1 ) ( 2x 3 5 2x) 4

+ 2x 3 5 2x 2, du bng xy ra 2x - 3 = 5 - 2x x = 2.

Vy: Phng trnh c nghim VT = VP = 2 x = 2.



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Bi 5

Gii cc bt phng trnh sau:

a/ x - 4 < x2 + x + 1

b/ + 2x 2x 1 1Gii

a/ Nu: x 4 PT x - 4 < x2 + x + 1 x2 > - 5 BPT c v s nghim

x 4(1)

Nu: x < 4 PT 4 - x < x2 + x + 1 x2 + 2x - 3 > 0 x > 1 x < -3

x < -3 1 < x < 4 (2)

T (1) v (2) x < -3 x > 1.

b/ + 2 2 x 1 1 x 2x 2x 1 1 (x 1) 1 x 1 1 x 1 1 x 0.

BTVN

Bi 1 Gii cc phng trnh: a/ x2 + 2x - 2 x + 1= 0 b/ x - 1 - x - 2 = 1

Bi 2Gii cc phng trnh sau:

a/ = 3 x 3x 5 b/ = 21 2x x 1

c/ + + = x 3 7 x 2x 8 (x = 5, x = 6) d/ + + =x 2 x 1 x 3 4 x 1 1

e/ + + = 3 x 2 x 1 2 x 2 x 1 f/ + + =x x(7 48) (7 48) 14

g/ + = +2x 94 96 x x 190x 9027 h/ + + = +26 x x 2 x 6x 13

Hng dn K: -2 x 6

VP = x2 - 6x + 13 = (x - 3)2 + 4 4. du bng xy ra x = 3.

VT2 = + + + + + + + 2 2 2 2( 6 x x 2) (6 x x 2)(1 1 ) ( 6 x x 2) 16

+ + 2( 6 x x 2) 4. Du bng xy ra = + =6 x x 2 x 4.

Vy: Phng trnh v nghim.



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Chuyn 7

Gii bi ton bng cch lp phng trnh, h phng trnh

I/ Cc bc gii mt bi ton bng cch lp phng trnh, H phng

trnh

B1: Lp phng trnh.- Chn n v xc nh iu kin cho n

- Biu th cc s liu cha bit qua n

- Tm mi lin h gia cc s liu lp phng trnh hoc h phng trnh

B2: Gii phng trnh hoc gii h phng trnh.

B3: Chn kt qu thch hp v tr li.

Ch :

- Qung ng = vn tc x thi gian (ton chuyn ng)- Sn lng = nng sut x thi gian (ton nng sut)

- Ngoi cch chn n trc tip i khi ta cn chn n gin tip c phng

trnh n gin hn.

II/ Bi tp.

*/ Ton chuyn ng

Bi 1

Mt ca n xui dng t A n B vi vn tc trung bnh l 30 km/h, sau li ngc t B n A. thi gian i xui t hn thi gian i ngc 40 pht.

Tnh di khc sng AB bit vn tc dng nc 3 km/h v vn tc ca ca

n khng i.Gii

Gi di khc sng AB l s (km)

thi gian ca n xui dng l: s/30 (gi)

thi gian ca n ngc dng l: s/(30 - 6) (gi)Theo bi ra ta c phng trnh:

= = = = =s s 2 s s 2 s s

2 2s 160 s 80(km)30 6 30 3 24 30 3 8 10

Bi 2

Mt ca n d nh i t A n B trong thi gian nh. Nu vn tc

ca n tng 3 km/h th n ni sm 2 gi. Nu vn tc ca n gim 3 km/h

th n ni chm 3 gi. Tnh chiu di khc sng.

Gii



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Gi vn tc d nh ca ca n l v (km/h) (v > 3), thi gian d nh l t (gi) (t

> 2), th chiu di khc sng AB l v.t (km)

Nu vn tc ca n tng 3 km/h th n ni sm 2 gi ta c: (v + 3)(y - 2) =

v.t

Nu vn tc ca n gim 3 km/h th n ni chm 3 gi ta c: (v-3)(y+3) =

v.t

Vy ta c h phng trnh:

+ = + = + = = + = + = = =

(v 3)(t 2) vt vt 2v 3t 6 vt 2v 3t 6 v 15

(v 3)(t 3) vt vt 3v 3t 9 vt 3v 3t 9 t 12

Vy khc sng AB di 12.15 = 180 (km).

Bi 3

Mt ca n xui khc sng di 40 km ri ngc khc sng y ht 4 gi ri.

Bit thi gian ca n xui 5 km bng thi gian ca n ngc 4 km. Tnh vn

tc dng nc.Gii

Gi vn tc ca dng nc l x (km.h) v vn t ca ca n l y (km/h),(x >y >0)

Do ca n xui khc sng di 40 km ri ngc khc sng y ht 4 gi ri ta c:

+ =+ 40 40 9

(1)x y x y 2

Do thi gian ca n xui 5 km bng thi gian ca n ngc 4 km ta c:

)2(45

yxyx =

+

T (1) v (2) ta c h phng trinh:

+ = + = = + + + = = = + + +

40 40 9 40 40 9 90 9x y x y 2 x y x y 2 x y 2

5 4 50 40 5 40 0 0x y x y x y x y x y x y

+ = = = =

x y 20 x 18

x y 16 y 2. Vy: vn tc dng nc l y = 2 km/h.

Bi 4

Mt ca n xui dng 45 km ri ngc dng 18 km. Bit rng thi gian xui

lu hn thi gian ngc l 1 gi v vn tc xui ln hn vn tc ngc l 6

km/h. tnh vn tc ca ca n lc ngc dng.Gii



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Gi v (km/h) l vn tc ca ca n lc ngc dng (v > 0) th thi gian xui dng 45

km l 45/(v+6) v thi gian ngc dng 18 km l 18/v. Theo bi ra ta c phng

trnh:

= = = + + =

+ =

2 2 v 1245 18 1 45v 18v 108 v 6v v 21v 108 0v 6 v v 9

.

Vy vn tc ca n lc ngc dng l: v = 12 hoc v = 9.

Bi 5*

Mt b na tri t do v mt ca n cng ri bn A xui dng sng. Ca

n xui dng c 96 km th tr v A, c i ln v mt 14 gi trn ng

v khi cn cch A 24 km th ca n gp b na tri. Tm vn tc ring ca

ca n v vn tc ca dng nc.

GiiGi vn tc ca ca n l x (km/h) v vn tc ca dng nc l y (km/h) (x>y>0)

Do Ca n xui dng c 96 km th tr v A, c i ln v mt 14 gi nn ta c:

+ =+ 96 96

14 (1)x y x y

Do trn ng v khi cn cch A 24 km th ca n gp b na tri nn ta c:

+ =+

96 72 24(2)

x y x y yT (1) v (2) ta c h phng trnh:

+ = + + = +

96 9614(1)

x y x y

96 72 24(2)

x y x y y

T (2) + = + = + + 96 72 24 4 3 1

x y x y y x y x y y

+ + = =2 2 2 2 '4xy 4y 3y 3xy x y x 7y (2)

T (1) + = + = = + + 2 2 '96 96 48 4814 7 96x 7(x y ) (1)

x y x y x y x y

Thay (2') vo (1') ta c: 96y = 48y2 y = 2 x = 14

Vy vn tc ring ca ca n l x = 14 v vn tc ca dng nc l y = 2.

Bi 6*

Mt chic tu thu xui t bn A n bn B ht 5 gi v ngc t bn Bv bn A ht 7 gi. Hi mt chic b c th tri theo dng nc th s i



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t bn A n bn B ht bao lu? Bit rng lt i cng nh lt v, tu

thu khng dng li ch no v vn gi nguyn vn tc ring ca n

(vn tc ring l vn tc khi nc yn lng).Gii

Gi khong cch AB l s Vn tc tu thu khi xui dng l: vx = s/5

Vn tc tu thu khi ngc dng l: vn = s/7

Ta c:

vx= vtu + vnc, vn = vtu - vnc vx- vn = 2vnc s/5 - s/7 = 2vnc vnc = s/35

Vy: Mt chic b tri t A n B ht 35 gi.

Bi 7

Qung ng AB gm mt on ln dc di 4 km v mt on xung dcdi 5 km. Mt ngi i xe p t A n B ht 40 pht v i t B v A ht

41 pht (vn tc ln dc lc i v v nh nhau, vn tc xung dc lc i

v v nh nhau). Tnh vn tc lc ln dc v lc xung dc.Gii

Gi vn tc lc ln dc l x (km/h), vn tc lc xung dc l y (km/h)

Theo bi ra ta c h phng trnh:

+ = + = = = = + = + = + =

4 5 40 20 25 200 9 36x y 60 x y 60 y 60 y 15

5 4 41 20 16 164 5 4 41 x 12x y 60 x y 60 x y 60

.

Vy: vn tc lc lndc l x = 12 km/h, vn tc lc xung xcl y = 15 km/h.

Bi 8

Mt ngi i xe p t A n B gm mt on ln dc AC v mt on

xung dc CB. Thi gian i AB l 2 gi, thi gian v BA l 1 gi 45 pht.

Tnh chiu di qung ng AB? Bit rng c ln dc th ngi i vi

vn tc 10 km/h v c xung dc th ngi i vi vn tc 15 km/h.Gii

Gi qung ng AB l s (km) ta c:

+ = + + + = + = + =

AC CB2

AC BC CB CA 15 s s 1510 15BC CA 7 10 15 4 10 15 410 15 4

+ = =6s 4s 225 s 22,5km



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Bi 9

Mt ngi i xe p t A n B ng di 78 km. Sau mt gi ngi th

hai i t B n A. hai ngi gp nhau ti C cch B l 36 km. Tnh thi

gian mi ngi i t lc khi hnh n lc gp nhau bit rng vn tc

ngi th hai ln hn vn tc ngi th nht l 4 km/h.Gii

Gi vn tc ngi th nht l v (km/h), (v > 0) th vn tc ngi th hai l v + 4.

Thi gian ngi th nht i l: 42/ v

Thi gian ngi th hai i l: 36/(v+4)

Theo bi ra ta c phng trnh:

= + = + = =+

2 242 36 1 6v 168 v 4v v 2v 168 0 v 14.v v 4

Vy: Thi gian ngi th nht i l: 42/ 14 = 3 gi

Thi gian ngi th hai i l: 36/ 18 = 2 gi.

Bi 10

Hai n v b i hai a im A v B cch nhau 39,5 km. Lc 6 gi

nv A i v pha B vi vn tc 6 km/h. Sau 2 gi n v B mi i

v pha A vi vn tc 5 km/h. Hi hai n v gp nhau lc my gi.Gii

Gi qung ng n v th nht i c cho n khi gp nhau l s1

Gi qung ng n v th hai i c cho n khi gp nhau l s2

Thi gian n v th nht i c cho n khi gp nhau l s1/6

Thi gian n v th hai i c cho n khi gp nhau l s2/5

Theo bi ra ta c phng trnh:

= = = = + = + = = + =

1 21 2 1 1

1 2 1 2 2

1 2

s s5s 6s 60 11s 297 s 272

6 56s 6s 237 6s 6s 237 s 12,5

s s 39,5

.

Vy: Thi gian n v th nht i c cho n khi gp nhau l 27/6 = 4,5 gi

hai i gp nhau lc 10gi 30 pht.

Bi 11*

Mt t ti i t A n B vi vn tc 30 km/h. Sau mt thi gian, mt

xe con cng xut pht t A vi vn tc 40km/h v nu khng c g thay

i th ui kp t ti ti B. Nhng sau khi i c na qung ng AB

th xe con tng vn tc ln thnh 45 km/h. nn sau 1 gi th uikp t ti. Tnh qung ng AB



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Gii

Gi qung ng AB l s (km)

Thi gian t ti i bnh thng l s/30 v thi gian xe con i bnh thng l s/40

Xe con xut pht sau t ti mt thi gian l: =s s s

30 40 120.

Qung ng m xe con i sau mt gi k t lc tng tc gp xe ti l 45 km

Nh vy thi gian m t ti i t A cho n khi gp xe con l: +s 45

2.30 30

Thi gian ng bng thi gian i ca xe con l: + +s s

12.40 120

.

Vy ta c phng trnh:

+ + = + + = =s s s 45 3s 2s 4s 120

1 s 1202.40 120 2.30 30 240 240 240 240

Vy: Qung ng AB = 120 km.

Bi 12*

Hai n v b i cng mt lc i t hai a im A v B gp

nhau. n v i t A mi gi i c 4 km. n v i t B mi gi i c

5 km. Mt ngi lin lc i xe p vi vn tc 12 km/h ln ng cng mt

lc vi cc n v b i, bt u t A gp n v i t B. Khi

gp n v ny ri, ngi lin lc lp tc quay v gpkj n v i t A

v khi gp n v ny ri li lp tcquay v gp n v i t B v

cnh th cho n khi hai n v gp nhau. Bit rngB di 27 km. Tnh

qung ng ngi lin lc i.Gii

Ta c thi gian m ngi lin lc chy i chy li ng bng thi gian m hai n

v b i gp nhau. Gi thi gian l t (gi).

Qung ng m nv i t A i c l: 4tQung ng m nv i t B i c l: 5t

Theo bi ra ta c: 4t + 5t = 27 t = 3.

Vy: Qung ng m ngi lin lc i l: 12.3 = 36 km

*/ Ton vi nc, ton nng sut ...

Bi 13

Ngi ta m ng thi hai vi nc cng chy vo mt b cn. Sau 4 gi b

y nc. Hi nu chy mt mnh, y b mi vi cn bao nhiu thigian? Bit rng lng nc chy ca vi th nht trong 2 gi 20 pht bng l-



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ng nc chy ca vi th hai trong 1 gi 45 pht.Gii

Gi t1 l thi gian vi mt chy mt mnh y b 1gi vi mt chy c 1/t1

b

Gi t2 l thi gian vi hai chy mt mnh y b 1gi vihi chy c 1/t2 b ta c: 4/t1 + 4/t2 = 1 (1)

Mt khc:

Trong 2 gi 20 pht = 7/3 gi vi mt chy c 7/3t1 b

Trong 1 gi 45 pht = 7/4 gi vi hai chy c 7/4t2 b

ta c: 7/3t1 = 7/4t2 (2)

T (1) v (2) ta c h phng trnh:

+ = + = = = = = = =

1 2 1 2 1 2 1 1 2 1

2 1 2 1 2

1 2

4 4 1t t 4t 4t t t 7t t t t 28/3

7 7 4t 3t 0 4t 3t 0 t 73t 4t

Vy: Vi mt chy mt mnh y b phi mt 28/3 gi

Vi hai chy mt mnh y b phi mt 7 gi.

Bi 14*

Ngi ta t mt vi nc chy vo mt b nc v mt vi nc chy ra lngchng b. Khi b cn, nu m c hai vi th sau 2 gi 42 pht b y n-

c. Cn nu ng vi chy ra, m vi chy vo th sau 1gi 30 pht y

b. Bit vi chy vo mnh gp 2 ln vi chy ra.

a/ Tnh thi gian nc chy vo t lc b cn n lc nc ngang ch t

vi chy ra.

b/ Nu chiu cao b l 2m th khong cch t ch t vi chy ra n

y b l bao nhiu.Gii

a/ Gi t (gi) l thi gian vi nc chy vo t khi b cn cho n khi mc nc

ngang ch t vi chy ra.

Trong 1 gi vi chy vo chy c 1/1,5 = 2/3 b

Trong 1 gi vi chy ra chy c 2/3 : 2 = 1/3 b

Nu m c hai vi th trong 1 gi lng nc chy vo b l: 2/3 - 1/3 = 1/3

Nhng trong t gi u ch c vi chy vo lm vic nn lng nc chy vo b l

2t/3 b



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Thi gian c hai vi lm vic l 2 gi 42 pht - t gi = (27/10 - t) gi l ng nc chy

vo b l (27/10 - t)/3 b

ta c phng trnh: + = + = =

27t2t 20t 27 10t 30 310 1 t

3 3 30 30 30 10.

Vy: thi gian nc chy vo t lc b cn n lc nc ngang ch t vi chy ra

l 0,3 gi.

b/ Nu chiu cao b l 2m th ring vi chy vo lm vic trong 1,5 gi th

mc nc cao 2 m ring vi chy vo lm vic trong 0,3 gi th mc nc cao

2.0,3/1,5 = 0,4 m

Vy: khong cch t ch t vichy ra n y b l 0,4 m.

Bi 15Mt phng hp c mt s dy gh, tng cng 40 ch. Do phi xp 55 ch

nn ngi ta k thm mt dy gh v mi dy xp thm mt ch. Hi lc

u c my dy gh trong phng?Gii

Gi s dy gh trong phng lc u l x (x nguyn dng) mi dy c 40/x ch

Lc sau c x + 1dy mi dy c 40/x + 1 gh ta c phng trnh:

+ + = + = = =240

(x 1)( 1) 55 x 14x 40 0 x 4 x 10x

vy lc u trong phng c 4 dy, mi dy 10 ch hoc c 10 dy mi dy 4 ch.

Bi 16

Mt x nghip d nh iu mt s xe chuyn 120 t hng. Nu

mi xe ch thm 1 t so vi d nh th s xe gim i 4 chic. Tnh s

xe d nh iu ng.Gii

Gi s xe d nh iu ng l x ( x nguyn dng) mi xe ch 120/x t. Theobi ra ta c phng trnh:

+ = = =

2120 1201 x 4x 480 0 x 24x x 4

Vy s xe d nh iu ng l 24 xe.

Bi 17

C hai i cng nhn, mi i phi sa 10 km ng. Thi gian i 1

lm nhiu hn i 2 l 1 ngy. Trong mt ngy, mi i lm c baonhiu km bit rng c hai i lm c 4,5 km trong mt ngy.



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Gii

Gi qung ng i 1 lm trong mt ngy l x (km), (0


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= + = =1x 150 450 x 200 150 (x 150) x 150 250.9 x 24009

Vy: S trngc tt c l 2400 v mi ngy bn c 400 qu.

BTVN

Bi 1Mt ngi i xe my t A n B vn tc 40 km/h. i c 15 pht ngi gp t

t B n vi vn tc 50 km/h. t n A ngh 15 pht ri tr v B v gp ngi

i xe my cch B 20 km. Tnh qung ng AB.

Hng dn

Gi C, D l ni m t gp ngi i xe my ln th 1 v ln th 2. Qung ng

CD l s (km).

ta c qung ng AC di 40.1/4 = 10 (km) thi gian ngi i xe my i t Cn D l s/40. Trong thi gian t i t C n A ri ngh 15 pht v i

on AD vi tng thi gian l (10+10+s)/50 + 1/4

Vy ta c phng trnh:+ + +

= + = + =s 10 10 s 1 5s 80 4s 50

s 13040 50 4 200 200 200

Vy: Qung ng AB di: 10 + 130 + 20 = 160.

Bi 2

Hai vi nc chy vo mt b th b s y trong 3 gi 20 pht. Ngi ta cho vith nht chy 3 gi, vith hai chy 2 gi th c hai vi chy c 4/5 b. Tnh

thi gian mi vi chy mt mnh y b.

Hng dn

Gi t1 l thi gian vi mt chy mt mnh y b 1gi vi mt chy c 1/t1

b

Gi t2 l thi gian vi hai chy mt mnh y b 1gi vihi chy c 1/t2 b

ta c: 10/3t1 + 10/3t2 = 1(1)

Mt khc:

Trong 3 gi vi mt chy c 3/t1 b

Trong 2 gi vi hai chy c 2/t2 b

ta c: 3/t1 + 2/t2 = 4/5 (2)

T (1) v (2) t1 = 5 gi, t2 = 10 gi.

Bi 3



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Trong mt lp hc nu b tr 4 em ngi mt gh th cn thiu mt gh. Nu b

tr 3 em ngi mt gh th cn tha 2 gh. Tnh s s lp v s gh ang c

trong lp.

Hng dn

Gi s s lp l s s/4 + 1 s/3 - 2 s/3 - s/4 = 3 s = 36 s gh = 10.Bi 4

Mt vn hnh ch nht c chu vi 450 m. Nu gim chiu di i 1/5 chiu di

c, tng chiu rng thm 1/4 chiu rng c th chu vi hnh ch nht khng

i. Tnh chiu di v chiu rng ca vng.

Bi 5

Mt sn hnh ch nht c di tch 720 m2. Nu tng chiu di 6m, gim chiu

rng 4m th din tch khng i. Tnh cc kch thc ca sn.



51/63


Chuyn 8

H thc lng trong tam gic

I/ L thuyt

1/ nh l Talt trong tam gic: DE // BC = =AD AE DE

AB AC BC

nh l Talt tng qut: AA1// BB1// CC1 = =1 1 1 1 1 1

AB BC ACA B B C A C

2/ Tnh cht ng phn gic ca tam gic:

ABC c AD l ng phn gic =DB ABDC AC

3/ Tam gic ng ging:

ABC A1B1C1= =

=

1

1

1

A AB B

C C

hoc

= =1 1 1 1 1 1

AB AC BCA B A C B C

hoc

= =

1

1 1 1 1

A A

AB ACA B A C

Nu: ABC A1B1C1 m =1 1

ABk

A Bth =

1 1

AHk

A Hv

=1 1 1

2ABC

A B C

Sk

S

4/ H thc lng trong tam gic vung:

-/ a2 = b2 + c2

-/ c2 = ac,, b2 = ab,

-/ h2 = b,c,

-/ ah = bc

-/ = +2 2 21 1 1h b c

II/ Bi tp

Bi 1

Cho tam gic ABC, ng thng d// BC ct AB ti M, ct AC ti N. Gi I, J

ln lt l trung im ca MN v BC.

a/ Chng minh rng: A, I, J thng hng.

b/ Gi P, Q, H ln lt l hnh chiu ca M, N, A ln BC, O = MP NQ, R l

trung im ca AH. Chng minh rng: J, O, R thng hng.

Gii


A

BD

C

A

B

C C1

B1

A1

A

CHB

c b

ac, b,

h

A

B E


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a/ p dng nh l Talt cho tam gic ABC ta c:

= = =MN AN MN /2 AN IN AN

BC AC BC /2 AC JC AC

A, I, J thng hng.

b/ Gi S l trung im ca PQ I, O, S thng hngv O l trung im ca IS, AH // IS theo cu a th ta c J, O, R thng hng.

Bi 2

Cho tam gic ABC vung ti A, phn gic trong AD, phn gic ngoi AE.

Cho bit AB < AC. Chng minh cc h thc sau:

a/ + =1 1 2AB AC AD

b/ =1 1 2AB AC AE

Gii

V DH AB, DK AC DH = DK =AD

2

a/ p dng nh l Talt cho ABC ta c:

= = = =DK CD 1 AD 1 1 1AB CB 2 2 AB2AB 2AD

= = = =HD CD 1 AD 1 1 1AC CB 2 2 AC

2AC 2AD

+ = =1 1 2 2

AB AC AD2AD

.

Cch khc:

Ch : SABC =2

1AB.ADsin(AB;AC)

a/ Ta c: SABC =1

2AB.AC = SABD + SACD =

1

2AB.ADsin450 +

1

2AC.ADsin450

2AB.AC = (AB+AC)AD

2 += =

+AB.AC 2AD AB AC 2

AB AC 2 AB.AC AD + =1 1 2

AB AC AD

b/ Ta c: SABC =1

2AB.AC = SAEC - SABE =

1

2AE.ACsin1350 -

1

2AB.AEsin450

AB.AC =AE

2(AC - AB) = =

AB.AC AE AC AB 2

AC AB AB.AC AE2

=1 1 2

AB AC AE.

Bi 3Cho tam gic ABC vung ti A, ngcao AH, trung tuyn AM. Chng minh


B

A

CED

KH

B J C

NIM

P QH

O

R

S

A


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cc h thc sau:

a/ =

2MH BM

2 1BH AB

b/ + = +2

2 2 2 BCAB AC 2AM2

Gii

a/ Do tam gic ABC vung ti A nn ta c:

= =2 2AB AB

BHBC 2BM

= = =

2 2 2AB 2BM ABMH MB BH BM

2BM 2BM

= = =

22 2 2 2

2 2

MH 2BM AB 2BM 2BM AB BM. 2 1

BH 2BM AB AB AB.

b/ Ta c: AB2

= AH2

+ HB2

, AC2

= AH2

+ HC2

AB2+ AC2 = 2AH2+ HB2+ HC2 = 2AH2+ (BM - HM)2+ (MC + HM)2 = 2AH2 + BM2+

MC2+2HM2- 2BM.HM + 2MC.HM = 2(AH2+ HM2) + (BC/2)2+ (BC/2)2 = 2AM2 +

BC2/2.

Bi 4

Cho tam gic u ABC, O l trung im ca BC, mt gc xOy = 60 0 c

cnh Ox, Oy lun ct AB, AC ti M v N.

a/ Chng minh rng OB2

= BM.CNb/ Chng minh rng tia MO, NO lun l phn gic ca gc BMN v CMN

c/ Chng minh rng ng thng MN lun tip xc vi mt ng trn c

nh khi gc xOy quay quanh O nhng hai cnh Ox, Oy vn ct hai cnh

AB v AC ca tam gic ABC.Gii

a/ Ta c: B = C = 600

O1 + O2 = 1200; O1 + M1 = 1200

M1= O2N1 = O1BOM CNO

BO/CN = BM/CO BO.CO = BM.CN BO2 = BM.CN

b/ T (a) ta c: = = =OM BM OM ON OM ONNO CO BM CO BM OB

Mt khc: MBO = MON = 600BOM ONM M1 = M2 OM l tia

phn gic ca BMN .

c/ Do O l giao im ca hai tia phn gic ca BMN v MNC O cch uAB, MN v AC.


B

A

CMH

A

B C

N

M

O

H


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Gi H l hnh chiu ca O ln AB OH = OB.sinB = =a 3 a 3.2 2 4

MN lun tip

xc vi ng trn c nh c tm O bn knh a 34

.

Bi 5

Cho tam gic u ABC, trn cc cnh BC, AB, AC ly ba im bt k O, M,

N sao cho O khc B, C v MON = 600.

Chng minh rng: BM.CN BC2/4. Du bng xy ra khi no?Gii

Ta c: BOM =1800 - B - BMO = 1200 - BMO

M: BOM = 1800 - MON - CON = 1200 - CON

BMO = CON BOM CNO

BM/CO = BO/CN BM.CN = BO.CO + =

2 2BO CO BC2 4

Bi 6

Cho tam gic ABC c ba gc nhn. Gi H l trc tm ca tam gic ABC, K

l chn ng cao v t A ca ABC. Chng minh rng:

2BC

KH.KA 4 .Gii

Xt AKB v CKH c: AKB = CKH = 900

BAK = HCK (hai gc nhn cnh tng ng vung gc)

AKB CKH =KA KCKB KH

+ = =

2 2KB KC BCKA.KH KB.KC

2 4

2BC

KH.KA4

Bi 7

Cho tam gic ABC vung ti A. Chng minh rng: =

+ABC AC

tg2 AB BC

Gii

a/ Xt ABD c A = 900

= =

AD ABC AD

tg ABD tgAB 2 AB


B

A

C

O

DE

A

B CK

H

A

B C

M

O

N


55/63


V ng phn gic BD ta c:

+= = = =+ +

DA BA DA DC DA DC ACDC BC BA BC AB BC AB BC

=

+

ABC ACtg

2 AB BC

.

Bi 8

Cho hnh thoi ABCD. Gi R1, R2 ln lt l bn knh cc ng trn ngoi

tip ABD v ABC. Gi a l di cnh hnh thoi.

a/ Chng minh rng: + =2 2 21 2

1 1 4R R a

.

b/ Tnh din tch hnh thoi theo R1 v R2.Gii

a/ Gi s trung trc cnh AB ct AC ti O1 v ct BD ti O2 O1 v O2 l tm cc

ng trn ngoi tip ABD v ABC O1A = R1 v O2B = R2.

O1AKABO = =1 1O A RAK aAB AO a 2AO

(1)

O2BKABO = =2 2O A RBK a

AB BO a 2BO(2)

T (1) v (2) = =

4 42 2

2 21 2

a a4AO , 4BOR R

( )

+ = + = + + =

2 2 4 2 42 2 2 2 2 2 21 2 1 2 1 2

1 1 1 1 1 1 44 AO BO a 4a a

R R R R R R a.

b/ Ta c: SABCD = 2OA.OB

AOB AKO2 = =2

2 2

OA AB ABOA

AK AO 2R

AOB O1KB = =2

1 1

OB AB ABOBKB O B 2R

=4

1 2

ABOA.OB4R R

Xt AOB ta c: AB2 = OA2 + OB2

= + = +

4 42 4

2 2 2 2

2 1 1 2

AB AB 1 1AB AB

4R 4R 4R 4R

+ = =

+

2 2 2 22 21 2 1 2

2 2 2 2

1 2 1 2

(R R ) 4R R1 AB AB

4R R R R.

Vy: = =+ +

4 4 3 31 2 1 2

ABCD2 2 2 2 2 21 2 1 2 1 2

16R R 8R R1OA.OB . S

4R R (R R ) (R R ).

Bi 9WWW.ToanTrungHocCoSo.ToanCapBa.Net Trang 55

C

B

A

D

K

O2

O1

O

a


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Chng minh rng trong mt tam gic bt k ta c:+ +< AB - BM

Xt ACM c: AM > AC - MC

Cng tng v ta c: 2AM > AB + AC - BC + >a

b c am

2.

Trn tia i ca tia MA ly im D sao cho MA = MD AB = CD

Xt ACD c: AD < AC + CD = AC + AB 2AM < AC + AB + AB + CD.

BTVN

Bi 1

Cho tam gic ABC, k ng cao AH, gi C1 l im i xng ca H qua AB, B1 lim i xng ca H qua AC. Gi giao im ca B1C1 vi AC v AB l I v K.

Chng minh rng ng BI, CK l ng cao ca tam gic ABC.

Bi 2

Cho tam gic ABC cn ti A v H l trung im ca cnh BC. Gi I l hnh chiu

vung gc ca H ln cnh AC v O l trung im ca HI. Chng minh rng hai

tam gic BIC v AOH ng dng vi nhau v AO vung gc vi BI.


A

D

CB

M

B

D

CA O


57/63


Chuyn 9

H thc lng trong ng trn

I/ L Thuyt

1/ nh ngha v s xc nh ng trn

*/ N. Tp hp cc im cch im O cho trc mt khong khng i R > 0 gil ng rn tm O bn knh R, k hiu: (O; R).

*/ Cho (O; R) v im M bt k. t d = OM ta c:

d < R M bn trong (O; R)

d = R M thuc (O; R)

d > R M bn ngoi (O; R)

*/ Hnh trn l tp hp cc im bn trong mt ng trn v cc imca

chnh ng trn .

M yhuc hnh trn (O; R) d R

*/ Cung trn l mt phn ca ng trn c gii hn bi hai im gi l mt ca

cung.

- on thng ni hai mt ca cung gi l dy trng cung

- Dy i qua tm gi l ng knh

- ng knh l dy cung ln nht ca ng trn

*/ Qu tch cc im M nhn on AB cho trc di mt gc vung l ng trn

tm I bn knh AB/2, k hiu (I; AB/2).

2/ V tr tng i gia ng thng v ng trn.

Cho ng trn (O; R) v ng thng , gi H l hnh chiu ca O ln ng

thng ta c:

OH > R khng ct ng trn (O; R)

OH = R ct ng trn (O; R) ti im H, gi l tip tuyn, H gi l tipim

OH < R ct ng trn (O; R) ti 2 im.

3/ V tr tng i gia hai ng trn.

Cho hai ng trn (O; R) v (O'; R')

OO' > R + R' Hai ng trn ngoi nhau

OO' = R + R' Hai ng trn tip xc ngoi

R - R'< OO' < R + R' Hai ng trn ct nhau



58/63


R - R'= OO' Hai ng trn tip xc trong

R - R'> OO' Hai ng trn ng nhau.

4/ Tip tuyn ca ng trn, dy cung ca ng trn

*/ Qua mt im nm trn ng trn c mt v ch mt tip tuyn vi ng

trn ng thng ny vung gc vi ng thng ni tm vi tip im.

*/ Qua mt im nm ngoi ng trn c hai tip tuyn vi ng trn ,

khong cch t im ti cc tip im th bng nhau.

*/ Trong mt ng trn hai dy cung bng nhau n cch u tm.

*/ Trong mt ng trn hai dy cung khc nhau, dy ln hn n gn tm

hn.

II/ Bi tp

Bi 1

Cho tam gic ABC v M l mt im thuc y BC v MD AB v ME

AC. Trn tia BD v CE ln lt ly cc im I, K sao cho D l trung im

ca BI, E l trung im ca CK.

a/ Chng minh rng bn im A, D, M, E cng thuc mt ng trn.

b/ Vi v tr no ca M trn y BC th 4 im B, I, K ,C cng thuc mt

ng trn.Gii

a/ Ta c D v E nhnon AM di mt gc

vung nn bn im A, D, M, E cng thuc

mt ng trn ng knh AM.

b/ MD l trung trc ca BI nn MB = MI

ME l trung trc ca CK nn MC = MK

4 im B, I, K ,C cng thuc mt ng

trn th phi c MB = MI = MK = MC M l trung im ca BC.Bi 2

Cho hnh vung ABCD, gi O l giao im ca hai ng cho AC v BD,

M l trung im ca o OA, N l trung im ca cnh BC. Chng minh

rng bn im C, M, N, D thuc mt ng trn v DN > MC.Gii

Gi H l trung im ca OC ta c NH OC

NH l ng trung bnh ca OBC NH = 12

OB


A

B CM

E

K

I

D

A

D C

N

B

M

H

O
http://var/www/apps/conversion/tmp/scratch_2/WWW.ToanTrungHocCoSo.ToanCapBa.Nethttp://var/www/apps/conversion/tmp/scratch_2/WWW.ToanTrungHocCoSo.ToanCapBa.Nethttp://var/www/apps/conversion/tmp/scratch_2/WWW.ToanTrungHocCoSo.ToanCapBa.Nethttp://var/www/apps/conversion/tmp/scratch_2/WWW.ToanTrungHocCoSo.ToanCapBa.Nethttp://var/www/apps/conversion/tmp/scratch_2/WWW.ToanTrungHocCoSo.ToanCapBa.Net


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v NH = OM =1

2OA (v OA = OB).

Vy OMD = HNM (v MH = OD =1

2AC

O = H = 900

, NH = OM) HMN = ODM DMN = 900

C, M nhn DN dimt gc vung bn im C, M, N, D thuc mt ng trn ng knh DN.

DN > MC v DN l ng knh cn MC l dy cung ca ng trn i qua

bn im C, M, N, D.

Bi 3

Cho ng trn tm O ng knh AB, im P di ng trn ng trn

sao cho PA < PB. Dng hnh vung APQR pha trong ng trn, tia PR

ct ng trn ti C.a/ Chng minh rng cung AC = cung CB

b/ Chng minh rng C l tm ng trn ngoi tip tam gic AQB

c/ Gi O' l tm ng trn ni tip tam gic APB. Chng minh rng O'

cng thuc ng trn qua A, Q, B.Gii

a/ Ta c: ABC = APC = 450 ( cng chn cung AC )

ACB vung ti C c B = 450

ACB cn ti C CA = CB cung AC = cung CB.

b/ Ta c: PC l ng trung trc ca AQ CA = CQ

kt hp vi cu (a) CA = CQ = CB C l tm

ng trn ngoi tip AQB.

c/ O' l tm ng trn ni tip APB O' thuc ng phn gic PC, v ng

phn gic BO' ta c: BO'C = O'BP + O'PB (gc ngoi ca

Các PP giai Toan lop 9

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