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

1D  2C

Câu 1: D

Câu 2: C

ko bit tu di ma lam suot ngay hoi bai

-buihoangvietlong bn quá đáng vừa thôi, ng ta ko biết mới phải hỏi, bn ko biết bn có phải hỏi ko? 

hoi lam gi lam mom   vao mang ma hoi neu ra moi trang nay thi chi xem rui h thoi

a) `x^4+2x^3-4x-4`

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

`=(x^2-2)(x^2+2)+2x(x^2-2)`

`=(x^2-2)(x^2+2+2x)`

b) `x^3-4x^2+12x-27`

`=(x^3-27)-(4x^2-12x)`

`=(x-3)(x^2+3x+9)-4x(x-3)`

`=(x-3)(x^2+3x+9-4x)`

`=(x-3)(x^2-x+9)`

c) `xy-4y-5x+20`

`=y(x-4)-5(x-4)`

`=(y-5)(x-4)`

a) Ta có: \(x^4+2x^3-4x-4\)

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

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

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

b) Ta có: \(x^3-4x^2+12x-27\)

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

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

c) Ta có: \(xy-4y-5x+20\)

\(=y\left(x-4\right)-5\left(x-4\right)\)

\(=\left(x-4\right)\left(y-5\right)\)

\(1,\\ 1,=15\left(x+y\right)\\ 2,=4\left(2x-3y\right)\\ 3,=x\left(y-1\right)\\ 4,=2x\left(2x-3\right)\\ 2,\\ 1,=\left(x+y\right)\left(2-5a\right)\\ 2,=\left(x-5\right)\left(a^2-3\right)\\ 3,=\left(a-b\right)\left(4x+6xy\right)=2x\left(2+3y\right)\left(a-b\right)\\ 4,=\left(x-1\right)\left(3x+5\right)\\ 3,\\ A=13\left(87+12+1\right)=13\cdot100=1300\\ B=\left(x-3\right)\left(2x+y\right)=\left(13-3\right)\left(26+4\right)=10\cdot30=300\\ 4,\\ 1,\Rightarrow\left(x-5\right)\left(x-2\right)=0\Rightarrow\left[{}\begin{matrix}x=2\\x=5\end{matrix}\right.\\ 2,\Rightarrow\left(x-7\right)\left(x+2\right)=0\Rightarrow\left[{}\begin{matrix}x=7\\x=-2\end{matrix}\right.\\ 3,\Rightarrow\left(3x-1\right)\left(x-4\right)=0\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{3}\\x=4\end{matrix}\right.\\ 4,\Rightarrow\left(2x+3\right)\left(2x-1\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=-\dfrac{3}{2}\\x=\dfrac{1}{2}\end{matrix}\right.\)

x³ + 2x²y + xy² - 4x

= x( x² + 2xy + y² - 4 )

= x[ ( x + y )² - 2² ]

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

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

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

\(=x\left(x+y\right)^2-4x=x\left[\left(x+y\right)^2-4\right]=x\left(x+y+2\right)\left(x+y-2\right)\)

a) x^ 3 + 2x^ 2y + xy ^2 − 4x

= x (x^ 2 + 2xy + y^ 2 − 4)

= x [(x + y)^ 2 − 2 ^2]

= x (x + y + 2)(x + y − 2) 

Lời giải:

a. 

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

$=xy(xy+1)-2(xy+1)=(xy+1)(xy-2)$

b. Bạn xem lại đoạn $-16x^2$ là dấu - hay + vậy?