a) Ta có: \(a^3y^3+125\)

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

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

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

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

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

\(=\left(3x+15\right)^2-\left(x-7\right)^2=\left(4x+8\right)\left(2x+22\right)=8\left(x+2\right)\left(x+11\right)\)

(x-7)^2 = x^2-14x+49 

<=> 9x^2+90x+225 -x^2+14x-49 
= 8x^2+104x+176 
= 8(x^2+13+22) 
<=> 8(x+2)(x+11)

9x²+90x+225-(x-7)²

=(3x+15)²-(x-7)²

=(3x+15-x+7)(3x+15+x-7)

=(2x+22)(4x+8)

=2 (x+11)4 (x+2)

=8 (x+11)(x+2)

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) Ta có: \(x^4y^4-z^4\)

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

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

b) Ta có: \(\left(x+y+z\right)^2-4z^2\)

\(=\left(x+y+z-2z\right)\left(x+y+z+2z\right)\)

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

a ,

b, 

c , 

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Cốc Cốc - Toán học | Trang chủ

    9x² + 90x + 225 - ( x - 7 )²

= ( 3x + 15 )² - ( x - 7 )²

= ( 3x + 15 + x - 7 )( 3x + 15 - x + 7 )

= ( 4x + 8 )( 2x + 22 )

= 4( x + 2 ) 2 ( x + 11 )

= 8( x + 2 )( x + 11 )

Hk tốt

9x²+90x+225-(x-7)²

=(3x)²+2.3x.15+15²-(x-7)²

=(3x+15)²-(x-7)²

=[(3x+15)-(x-7)][(3x+15)+(x-7)]

=(3x+15-x+7)(3x+15+x-7)

=(2x+22)(4x+8)

=2(x+11).4(x+2)

=8(x+11)(x+2)

dấu ^ là mũ nha mn