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a) x2 + 6x + 9 = x2 + 2 . x . 3 + 32 = (x + 3)2
b) 10x – 25 – x2 = -(-10x + 25 +x2) = -(25 – 10x + x2)
= -(52 – 2 . 5 . x – x2) = -(5 – x)2
c) 8x3 - 1/8 = (2x)3 – (1/2)3 = (2x - 1/2)[(2x)2 + 2x . 12 + (1/2)2]
= (2x - 1/2)(4x2 + x + 1/4)
d)1/25x2 – 64y2 = (1/5x)2(1/5x)2- (8y)2 = (1/5x + 8y)(1/5x - 8y)
a) x2 + 6x + 9 = x2 + 2 . x . 3 + 32 = (x + 3)2
b) 10x – 25 – x2 = -(-10x + 25 +x2) = -(25 – 10x + x2)
= -(52 – 2 . 5 . x – x2) = -(5 – x)2
c) 8x3 - 1/8 = (2x)3 – (1/2)3 = (2x - 1/2)[(2x)2 + 2x . 12 + (1/2)2]
= (2x - 1/2)(4x2 + x + 1/4)
d)1/25x2 – 64y2 = (1/5x)2(1/5x)2- (8y)2 = (1/5x + 8y)(1/5x - 8y)
1. \(x\left(x^2-5xy-14y^2\right)=x\left(x^2-7xy+2xy-14y^2\right)\)
\(=x\left(x-2\right)\left(x-7\right)\)
2. \(x^4+2x^2+1-9x^2=\left(x^2+1\right)^2-\left(3x\right)^2=\left(x^2+1-3x\right)\left(x^2+1+3x\right)\)
3. \(4x^4+4x^2+1-16x^2=\left(2x^2+1\right)^2-\left(4x\right)^2=\left(2x^2-4x+1\right)\left(2x^2+4x+1\right)\)
4. \(x^2+x+7x+7=\left(x+7\right)\left(x+1\right)\)
5. \(x\left(x^2-5x-14\right)=x\left(x^2-7x+2x-14\right)=x\left(x+2\right)\left(x-7\right)\)
Phân tích đa thức thành nhân tử :
1.x3-5x2y-14xy2
2.x4-7x2+1
3.4x4-12x2+1
4.x2+8x+7
5.x3-5x2-14x
c, =(5x)^3 + (y^2)^ 3 = (5x+y^2)(25x^2 - 5xy^2 + y^4)
d, = (0,5.(a+1))^3-1^3 = ( 0,5(a+1) - 1 ) ( 0,25(a+1) ^2 +a,5(a+1) + 1)
e,2x( x+ 1 ) + 2(x+ 1 ) = 2(x+1)(x+1) = 2(x+1)^2
g, y^2 (x^2 + y) - zx^2 - zy = x^2.y^2 - z.x^2 + y^3 - zy = x^2 (y^2 - z) + y (y^2 -z) = (x^2 +y) (y^2 -z)
h,4.x(x-2y) + 8.y(2y -x) = 4x( x- 2 y ) -8 (x - 2y) = (4x - 8) (x-2y)=4(x-2)(x-2y)
k,=(x+1)(3x(x+1)-5x+7) =(x+1) (3x^2 +3x - 5x + 7)
a)
\(14x^2y-21xy^2+28x^2y^2\)
\(=7xy(2x-3y+4xy)\)
b) \(x(x+y)-5x-5y=x(x+y)-5(x+y)=(x-5)(x+y)\)
c)
\(10x(x-y)-8(y-x)=10x(x-y)+8(x-y)\)
\(=(x-y)(10x+8)=2(x-y)(5x+4)\)
a. \(14x^2y-21xy^2+28x^2y^2\)
\(=7xy\left(2x-3y+4xy\right)\)
b. \(x\left(x+y\right)-5x-5y\)
\(=x\left(x+y\right)-5\left(x+y\right)\)
\(=\left(x-5\right)\left(x+y\right)\)
c. \(10x\left(x-y\right)-8\left(y-x\right)\)
\(=10x\left(x-y\right)+8\left(x-y\right)\)
\(=\left(10x+8\right)\left(x-y\right)\)
d. \(\left(3x+1\right)^2-\left(x+1\right)^2\)
\(=\left(3x+1+x+1\right)\left(3x+1-x-1\right)\)
\(=2x\left(4x+2\right)\)
\(=4x\left(2x+1\right)\)
e. Vì bài này giải không ra nên mình nghĩ nó sai đề, sửa lại tí nhé!
\(x^3+y^3+z^3-3xyz\)
\(=\left(x+y\right)^3+z^3-3xy\left(x+y\right)-3xyz\)
\(=\left(x+y+z\right)\left[\left(x+y\right)^2-z\left(x+y\right)+z^2\right]-3xy\left(x+y+z\right)\)
\(=\left(x+y+z\right)\left(x^2+2xy+y^2-xz+zy+z^2-3xy\right)\)
g. \(5x^2-10xy+5y^2-20z^2\)
\(=5\left(x^2-2xy+y^2-4z^2\right)\)
\(=5\left[\left(x-y^2\right)-4z^2\right]\)
\(=5\left(x-y+z\right)\left(x-y-z\right)\)
h. \(x^3-x+3x^2y+3xy^3+y^3-y\)
\(=\left(x^3+3x^2y+3xy^2+y^2\right)-\left(x+y\right)\)
\(=\left(x+y\right)^3-\left(x+y\right)\)
\(=\left(x+y\right)\left[\left(x+y\right)^2-1\right]\)
\(=\left(x+y\right)\left(x+y-1\right)\left(x+y+1\right)\)
i. \(x^2+7x-8\)
\(=x^2-x+8x-8\)
\(=x\left(x-1\right)+8\left(x-1\right)\)
\(=\left(x+8\right)\left(x-1\right)\)
a) \(3x\left(x+1\right)^2-5x^2\left(x+1\right)+7\left(x+1\right)\)
\(=\left(x+1\right)\left[3x\left(x+1\right)-5x^2+7\right]\)
\(=\left(x+1\right)\left(3x^2+3x-5x^2+7\right)\)
\(=\left(x+1\right)\left(-2x^2+3x+7\right)\)
\(=-\left(x+1\right)\left(2x^2-3x-7\right)\)
b) \(\left(x+y\right)\left(2x-y\right)-\left(3x-y\right)\left(y-2x\right)\)
\(=\left(x+y\right)\left(2x-y\right)+\left(3x-y\right)\left(2x-y\right)\)
\(=\left(2x-y\right)\left(x+y+3x-y\right)\)
\(=4x\left(2x-y\right)\)
c) \(5u\left(u-v\right)^2+10u^2\left(v-u\right)^2\)
\(=5u\left(u-v\right)^2+10u^2\left(u-v\right)^2\)
\(=5u\left(u-v\right)^2\left(1+2u\right)\)
Trả lời:
a, 3x ( x + 1 )2 - 5x2 ( x + 1 ) + 7 ( x + 1 )
= ( x + 1 )[ 3x ( x + 1 ) - 5x2 + 7 ]
= ( x + 1 )( 3x2 + 3x - 5x2 + 7 )
= ( x + 1 )( - 2x2 + 3x + 7 )
b, ( x + y )( 2x - y ) - ( 3x - y )( y - 2x )
= ( x + y )( 2x - y ) + ( 3x - y )( 2x - y )
= ( 2x - y )( x + y + 3x - y )
= 4x ( 2x - y )
c, 5u ( u - v )2 + 10u2 ( v - u )2
= 5u ( u - v )2 + 10u2 ( u - v )2
= 5u ( u - v )2( 1 + 2u )
1/a ) = (x+y)3 -(x+y)
= (x+y)[(x+y)2+1]
c) = 5(x2-xy+y2)-20z2
=5(x-y)2-20z2
= 5 [ (x-y)2- 4z2 ]
=5(x-y-4z)(x-y+4z)
Bài 1:
a) x3-x+3x2y+3xy2+y3-y
=x3+2x2y-x2+xy2-xy+x2y+2xy2-xy+y3-y2+x2+2xy-x+y2-y
=x(x2+2xy-x+y2-y)+y(x2+2xy-x+y2-y)+(x2+2xy-x+y2-y)
=(x2+2xy-x+y2-y)(x+y+1)
=[x(x+y-1)+y(x+y-1)](x+y+1)
=(x+y-1)(x+y)(x+y+1)
c) 5x2-10xy+5y2-20z2
=-5(2xy-y2+4z2-2)
Bài 2:
5x(x-1)=x-1
=>5x2-6x+1=0
=>5x2-x-5x+1
=>x(5x-1)-(5x-1)
=>(x-1)(5x-1)=0
=>x=1 hoặc x=1/5
b) 2(x+5)-x2-5x=0
=>2(x+5)-x(x+5)=0
=>(2-x)(x+5)=0
=>x=2 hoặc x=-5