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

x³-3x²-4x+12=(x³-3x²)-(4x-12)=x²(x-3)-4(x-3)=(x-3)(x²-4)=(x-3)(x-2)(x+2)

\(x^3-3x^2-4x+12=\left(x^3-3x^2\right)-\left(4x-12\right)\\ =x^2\left(x-3\right)-4\left(x-3\right)=\left(x-3\right)\left(x^2-4\right)\\ =\left(x-3\right)\left(x-2\right)\left(x+2\right)\)

\(x^3-3x^2-4x+12\)

\(=\left(x^3-3x^2\right)-\left(4x-12\right)\)

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

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

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

\(x^4-4x^3-2x^2-3x+2\)

\(\Leftrightarrow x^4+x^3-5x^3+x^2-5x^2+2x^2-5x+2x+2\)

\(\Leftrightarrow x^4+x^3+x^2-5x^3-5x^2-5x+2x^2+2x+2\)

\(\Leftrightarrow x^2\left(x^2+x+1\right)-5x\left(x^2+x+1\right)+2\left(x^2+x+1\right)\)

\(\Leftrightarrow\left(x^2-5x+2\right)\left(x^2+x+1\right)\)

Xin tick ạ !!!

\(x^3-3x^2+4x-2=x^3-x^2-2x^2+2x-2=x^2\left(x-1\right)-2x\left(x-1\right)+2\left(x-1\right)\)

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

x³ - 3x² + 4x - 2

= x³ - x² - 2x² + 2x + 2x - 2

= x²( x -1 ) - 2x( x - 1 ) + 2( x - 1 )

= ( x - 1 )( x² - 2x + 2 )

a: \(2y\left(x+2\right)-3x-6\)

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

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

b: \(3\left(x+4\right)-x^2-4x\)

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

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

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

c: \(2\left(x+5\right)-x^2-4x\)

\(=2x+10-x^2-4x\)

\(=-x^2-2x+10\)

\(=-x^2-2x-1+11\)

\(=11-\left(x^2+2x+1\right)\)

\(=11-\left(x+1\right)^2\)

\(=\left(\sqrt{11}-x-1\right)\left(\sqrt{11}+x+1\right)\)

d: \(x^2+6x-3x-18\)

\(=\left(x^2+6x\right)-\left(3x+18\right)\)

\(=x\left(x+6\right)-3\left(x+6\right)\)

\(=\left(x+6\right)\left(x-3\right)\)

Làm theo kiểu PP số 7 nhé bạn 

\(x^3-3x^2+4x-2\)

\(=x^3-x^2-2x^2+2x+2x-2\)

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

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

cái này hay nha

nhưng mik chỉ bik đáp án

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

phân tích như nào ạ ? TT^TT

e) \(=x^2\left(x+1\right)-2x\left(x+1\right)+3\left(x+1\right)=\left(x+1\right)\left(x^2-2x+3\right)\)

g) \(=x^2\left(3x-1\right)-x\left(3x-1\right)+4\left(3x-1\right)=\left(3x-1\right)\left(x^2-x+4\right)\)

h) \(=3x^2\left(2x+1\right)-x\left(2x+1\right)+\left(2x+1\right)=\left(2x+1\right)\left(3x^2-x+1\right)\)

i) \(=2x^2\left(2x+1\right)+2x\left(2x+1\right)+\left(2x+1\right)=\left(2x+1\right)\left(2x^2+2x+1\right)\)