a) Ta có: \(a^3y^3+125\)

\(=\left(ay+5\right)\left(a^2y^2-5ay+25\right)\)

b) Ta có: \(8x^3-y^3-6xy\cdot\left(2x-y\right)\)

\(=\left(2x-y\right)\left(4x^2+2xy+y^2\right)-6xy\left(2x-y\right)\)

\(=\left(2x-y\right)\left(4x^2+2xy-6xy+y^2\right)\)

\(=\left(2x-y\right)^3\)

16x^4_y2-25a²b²

1) \(x^6+1\)

\(=x^6+x^4-x^4+x^2-x^2+1\)

\(=\left(x^6-x^4+x^2\right)+\left(x^4-x^2+1\right)\)

\(=x^2\left(x^4-x^2+1\right)+\left(x^4-x^2+1\right)\)

\(=\left(x^2+1\right)\left(x^4-x^2+1\right)\)

2) \(x^6-y^6\)

\(=\left(x^3+y^3\right)\left(x^3-y^3\right)\)

\(=\left(x+y\right)\left(x^2-xy+y^2\right)\left(x-y\right)\left(x^2+xy+y^2\right)\)

Lời giải:
$x^4y^4-z^4=(x^2y^2)^2-(z^2)^2=(x^2y^2-z^2)(x^2y^2+z^2)$

$=(xy-z)(xy+z)(x^2y^2+z^2)$

$(x+y+z)^2-4z^2=(x+y+z)^2-(2z)^2=(x+y+z-2z)(x+y+z+2z)$
$=(x+y-z)(x+y+3z)$

$\frac{-1}{9}x^2+\frac{1}{3}xy-\frac{1}{4}y^2=\frac{-4x^2+12xy-9y^2}{36}$

$=-\frac{4x^2-12xy+9y^2}{36}=-\frac{(2x-3y)^2}{36}=-\left(\frac{2x-3y}{6}\right)^2$

Câu trả lời của cô quá đúng luôn đấy

a)8x²+12x²y+6xy²+y³

<=> ( 2x + y)³

a) (2x+y)³

c)(x²-y²)(x⁴+x²y²+y⁴)

d)-x³+9x²-27x+27

<=> -(x³-9x²+27x-27)

<=>-(x-3)³

a) ( 3x - 1 )² - 4 

= ( 3x - 1 ) - 2²

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

= ( 3x - 3 )( 3x + 1 )

= 3( x - 1 )( 3x + 1 )

b) ( x + y )² - x²

= ( x + y - x )( x + y + x )

= y( 2x + y )

c) 100 - ( 2x - y )²

= 10² - ( 2x - y )²

= [ 10 - ( 2x - y ) ][ 10 + ( 2x - y ) ]

= ( 10 - 2x + y )( 10 + 2x - y )

d) ( 2x - 1 )² - ( x - 1 )²

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

= ( 2x - 1 - x + 1 )( 2x - 1 + x - 1 )

= x( 3x - 2 )

e) 4( x + 6 )² - 9( 1 + x )²

= 2²( x + 6 )² - 3²( 1 + x )²

= ( 2x + 12 )² - ( 3 + 3x )²

= [ ( 2x + 12 ) - ( 3 + 3x ) ][ ( 2x + 12 + ( 3 + 3x ) ]

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

= ( 9 - x )( 5x + 15 )

= 5( 9 - x )( x + 3 )

a) =( x²+y²-5)²-[2(xy-2)]²

⁼( x²+y²-5)²- (2xy-4)²

=(x²+y²-5+2xy-4)(x²+y²-5-2xy+4)

=[(x+y)²-9][(x-y)²-1]

phân tích tiếp HĐT 2 ở 2 thừa số