Phân tích các đa thức sau thành nhân tử chung bằng cách dùng hằng đẳng thức:
a, 10x ( x – y) -8y (y -x)
b, 2x (x + 3) + 2 (x + 3)
c, 4x ( x – 2y) + 8y (2y – x)
d, ( x + y) ³ – x ³ – y ³
e, y ² ( x ² + y) – zx ² – zy
f, 1/2x ( x ² – 4) + 4 ( x + 2)
g, 4x ³y – 9xy
a, 10x ( x – y) -8y (y -x)
= 10x(x – y) + 8y(x – y)
= (x – y)(10x + 8y)
= 2(x – y)(x + y)
b, 2x (x + 3) + 2 (x + 3)
= (x + 3)(2x + 2)
= 2(x + 1)(x + 3)
c, 4x ( x – 2y) + 8y (2y – x)
= 4x(x – 2y) – 8y(x – 2y)
= (x – 2y)(4x – 8y)
= 4(x – 2y)(x – 2y)
= 4(x – 2y)²
d, ( x + y)³ – x ³ – y ³
= (x + y)³ – (x³ + y³)
= (x + y)³ – (x + y)(x² – xy + y²)
= (x + y)[(x + y)² – (x² – xy + y²)]
= (x + y)(x² + 2xy + y² – x² + xy – y²
= 3xy(x + y)
e, y ² ( x ² + y) – zx ² – zy
y²(x² + y) – z(x² + y)
= (x² + y)(y² – z)
f, 1/2x ( x ² – 4) + 4 ( x + 2)
= $\frac{1}{2}$x(x – 2)(x + 2) + 4(x + 2)
= (x + 2)[ $\frac{1}{2}$x(x – 2) + 4]
= (x + 2)( $\frac{1}{2}$x² – 1 + 4)
= (x + 2)( $\frac{1}{2}$x² – 3)
g, 4x ³y – 9xy
= xy(4x² – 9)
= xy[(2x)² – 3²]
= xy(2x – 3)(2x + 3)
a, Ta có :
$10x ( x – y) -8y (y -x)$
$= 10x(x – y) + 8y(x – y)$
$= (x – y)(10x + 8y)$
$= 2(x – y)(x + y)$
b,Ta có :
$2x (x + 3) + 2 (x + 3)$
$= (x + 3)(2x + 2)$
$= 2(x + 1)(x + 3)$
c, Ta có :
$4x ( x – 2y) + 8y (2y – x)$
$= 4x(x – 2y) – 8y(x – 2y)$
$= (x – 2y)(4x – 8y)$
$= 4(x – 2y)(x – 2y)$
$= 4(x – 2y)^2$
d,Ta có :
$( x + y)^3 – x^3 – y^3$
$= (x + y)^3 – (x^2 + y^3)$
$= (x + y)^3 – (x + y)(x^2 – xy + y^2)$
$= (x + y)[(x + y)^2 – (x^2 – xy + y^2)]$
$= (x + y)(x^2 + 2xy + y^2 – x^2 + xy – y^2$
$= 3xy(x + y)$
e, Ta có :
$y ^2 ( x^2 + y) – zx ^2 – zy$
$y^2(x^2+ y) – z(x^2 + y)$
$= (x^2 + y)(y^2 – z)$
f, Ta có :
$1/2x ( x ^2 – 4) + 4 ( x + 2)$
$= 12x(x – 2)(x + 2) + 4(x + 2)$
$= (x + 2)[ 12x(x – 2) + 4]$
$= (x + 2)( 12x^2 – 1 + 4)$
$= (x + 2)( 12x^2 – 3)$
g, Ta có
$4x ^3y – 9xy$
$= xy(4x^2 – 9)$
$= xy[(2x)^2 – 3^2]$
$= xy(2x – 3)(2x + 3)$