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= (2x)3- y3 + ( 4x2y +2xy + y2)
= (2x - y +1)(4x2y + 2xy + y2)
( hiu k? a3 - b3 mà a = 2x ; b =y)
a, ( 2x - 1)^2 - (4x + 2) ^2 = ( 2x - 1 - 4x- 2) ( 2x - 1 + 4x + 2) = (-2x-3)(6x+1) = - (2x+3)(6x+1)
b, 8x^3 + 12x^2y + 6xy^2 + y^3
= (2x)^2 + 3.(2x)^2 . y + 3.2x.y^2 + y^3
= (2x + y)^3
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)
\(a,4x^4-8x^3+4x^2\)
\(=4x^2\cdot\left(x^2-2x+1\right)\)
\(=4x^2\cdot\left(x-1\right)^2\)
\(b,x^2-y^2+5\cdot\left(y-x\right)\)
\(=\left(x-y\right)\cdot\left(x+y\right)-5\cdot\left(x-y\right)\)
\(=\left(x-y\right)\cdot\left(x+y-5\right)\)
\(c,3x^2-6xy+3y^2-12z^2\)
\(=3\cdot\left(x^2-2xy+y^2-4x^2\right)\)
\(=3\cdot\left[\left(x-y\right)^2-\left(2x\right)^2\right]\)
\(=3\cdot\left(x-y-2x\right)\cdot\left(x-y+2x\right)\)
a) \(4x^2-8x+4-9\left(x-y\right)^2\)
\(=4\left(x^2-2x+1\right)-9\left(x-y\right)^2\)
\(=\left[2\left(x-1\right)\right]^2-\left[3\left(x-y\right)\right]^2\)
\(=\left(2x-2+3x-3y\right)\left(2x-2-3x+3y\right)\)
\(=\left(5x-3y-2\right)\left(3y-x-2\right)\)
b) \(x^3-4x^2+12x-27\)
\(=\left(x^3-27\right)-\left(4x^2-12x\right)\)
\(=\left(x-3\right)\left(x^2+3x+9\right)-4x\left(x-3\right)\)
\(=\left(x-3\right)\left(x^2-x+9\right)\)
tớ lm câu a thui nha , tại khó quá ^^
a/ \(=3x^6+3x^5+6x^4+3x^3+3x^2-7x^5-7x^4-14x^3-7x^2-7x+3x^4+3x^3+6x^2+3x+1\)
\(=3x^2\left(x^4+x^3+2x^2+x+1\right)-7x\left(x^4+x^3+2x^2+x+1\right)+3\left(x^4+x^3+2x^2+x+1\right)\)
\(=\left(3x^2-7x+3\right)\left(x^4+x^3+x^2+x^2+x+1\right)\)
\(=\left(3x^2-7x+3\right)\left[x^2\left(x^2+x+1\right)+\left(x^2+x+1\right)\right]\)
\(=\left(3x^2-7x+3\right)\left(x^2+1\right)\left(x^2+x+1\right)\)
A) 1/2 x(x^2-4)+4(x+2)
=1/2x(x-2)(x+2)+4(x+2)
=(x+2)(1/2x^2-x+4)
b) 21(x-y)^2-7(x-y)^3
= (x-y)^2(21-7x+7y)
=(x-y)^2.7(3-x+y)
c) 1/8x^3-3/4x^2+3/2x-1
=(1/2x)^3-3.(1/2x)^2.1+3.1/2x.1^2-1
=(1/2x-1)^3
x6+3x4y2-8x3y3+3x2y4+y6= x6+3x4y2+3x2y4+y6-8x3y3=(x2+y2)3-(2xy)3
= (x2+y2-2xy)[(x2+y2)2+2xy(x2+y2)+(2xy)2]= (x-y)2(x4+6x2y2+y4+2x3y+2xy3)
(x2+y2-5)2-4x2y2-16xy-16=(x2+y2-5)2-(4x2y2+16xy+16)=(x2+y2-5)2-(2xy+4)2
=(x2+y2-5+2xy+4)(x2+y2-5-2xy-4)=(x2+2xy+y2-1)(x2-2xy+y2-9)=[(x+y)2-1][(x-y)2-32]=(x+y-1)(x+y+1)(x-y-3)(x-y+3)
x4+324=x4+36x2+324-36x2=(x2+18)2-(6x)2=(x2+18-6x)(x2+18+6x)
Phân tích các đa thức sau thành nhân tử :
a) x - y + 5x - 5y
= ( x + 5x ) - ( y + 5y )
= x . ( 1 + 6 ) - y . ( 1 + 6 )
= ( 1 + 6 ) . ( x - y )
\(a,x-y+5x-5y=\left(x-y\right)+5\left(x-y\right)=6\left(x-y\right)\)
a,x4-4x3+8x2-16x+16
=(x4-4x3+4x2)+(4x2-16x+16)
=(x^2-2x)^2+(2x-4)^2
=x^2(x-2)^2+4(x-2)^2
=(x-2)^2(x^2+4)
\(8x^3+4x^2-y^3-y^2\)
\(=\left(8x^3-y^3\right)+\left(4x^2-y^2\right)\)
\(=\left(2x-y\right)\left(4x^2+2xy+y^2\right)+\left(2x-y\right)\left(2x+y\right)\)
\(=\left(2x-y\right)\left(4x^2+2xy+y^2+2x+y\right)\)