a: \(=x\left(x^2+4x+4-z^2\right)\)

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

z còn câu b

2x2 + 4x + 2 – 2y2 (có nhân tử chung là 2)

= 2.(x2 + 2x + 1 – y2) (Xuất hiện x2 + 2x + 1 là hằng đẳng thức)

= 2[(x2 + 2x + 1) – y2]

= 2[(x + 1)2 – y2] (Xuất hiện hằng đẳng thức (3))

= 2(x + 1 – y)(x + 1 + y)

\(1,=x\left(x^2-2x+1-y^2\right)=x\left[\left(x-1\right)^2-y^2\right]=x\left(x-y-1\right)\left(x+y-1\right)\\ 2,=\left(x+y\right)^3\\ 3,=\left(2y-z\right)\left(4x+7y\right)\\ 4,=\left(x+2\right)^2\\ 5,Sửa:x\left(x-2\right)-x+2=0\\ \Leftrightarrow\left(x-2\right)\left(x-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)

Lời giải:
$2x^2+3xy-5y^2=(2x^2-2xy)+(5xy-5y^2)$

$=2x(x-y)+5y(x-y)=(x-y)(2x+5y)$

\(\left(x^2+4x+8\right)^2+3x\left(x^2+4x+8\right)+2x^2=\left(x^2+4x+8+\dfrac{3}{2}x\right)^2-\dfrac{1}{4}x^2=\left(x^2+\dfrac{11}{2}x+8\right)^2-\left(\dfrac{1}{2}x\right)^2=\left(x^2+\dfrac{11}{2}x+8-\dfrac{1}{2}x\right)\left(x^2+\dfrac{11}{2}x+8+\dfrac{1}{2}x\right)=\left(x^2+5x+8\right)\left(x^2+6x+8\right)=\left(x+2\right)\left(x+4\right)\left(x^2+5x+8\right)\)

\(\left(x^2+4x+8\right)^2+3x\left(x^2+4x+8\right)+2x^2\)

\(=\left(x^2+4x+8\right)^2+x\left(x^2+4x+8\right)+2x\left(x^2+4x+8\right)+2x^2\)

\(=\left(x^2+4x+8\right)\left(x^2+5x+8\right)+2x\left(x^2+5x+8\right)\)

\(=\left(x^2+5x+8\right)\left(x+2\right)\left(x+4\right)\)

\(=\left(y+4\right)^2-9x^2=\left(y-3x+4\right)\left(y+3x+4\right)\)

16 - 9y^2 + y^2 + 8y

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

= ( 4 + y + 3x ) ( 4 + y - 3x )

a. \(x^2\) - 9y²

= (\(x\))² - (3y)²

= (\(x\) - 3y)(\(x\) + 3y)

x² - 9y² = x² - (3y)²= (x - 3y)(x + 3y)

b: \(=\left(x-5\right)^2-9y^2\)

\(=\left(x-5-3y\right)\left(x-5+3y\right)\)