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: \(\left(x+y+z\right)^2-4z^2\)

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

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

=4^2.3-4y^2-4y

=4.(12-y^2-y)

hol tot 

nho k nhe ae

good luck

\(48-4y^2-4y\)

\(=-\left(4y^2+4y-48\right)\)

\(=-\left(4y^2+4y+1-49\right)\)

\(=-\left[\left(2y+1\right)^2-7^2\right]\)

\(=-\left(2y+1-7\right)\left(2y+1+7\right)\)

\(=-\left(2y-6\right)\left(2y+8\right)\)

\(=-4\left(y-3\right)\left(y+4\right)\)

\(=x^2+2\cdot x\cdot2y+\left(2y\right)^2=\left(x+2y\right)^2\)

\(x^2+4xy+4y^2=\left(x+2y\right)^2\)(Nhớ k cho mình với nhé!)

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

\(x^2+4xy+4y^2=\left(x+2y\right)^2\)²

k mình nha

=(x^2-4x+4)-4y^2

=(x-2)^2-(2y)^2

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

nếu thấy đúng thì k nhe

mk k lại cho

Câu 1:8-x^3=2^3-x^3=(2-x)(4+2x+x^2)

Câu 2:Ta có:x^2-5x+4

=(x^2-2x5/2+25/4)-9/4

=(x-5/2)^2-(3/2)^2

=(x-5/2-3/2)(x-5/2+3/2)

=(x-4)(x-1)

->đa thức B là:(x-4)

->hệ số tự do của đa thức B là:-4

\(48-4y^2-4y=-\left(4y^2+4y-48\right)\)

                 \(=-\left[\left(2y\right)^2+2.2y+1-49\right]\)

         \(=-\left[\left(2y+1\right)^2-7^2\right]\)

\(=-\left(2y-6\right)\left(2y+8\right)\)