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

(3x+2)²-(x-6)²=(3x+2-x+6)(3x+2+x-6)=(2x+8)(4x-4)=8(x+4)(x-1)

\((3x+2)^2-(x-6)^2=(3x+2-x+6)(3x+2+x-6) =(2x+8)(4x-4)=2.4(x+4)(x-1)=8(x+4)(x-1)\)

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

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

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

\(=8\left(x-1\right)\left(x+4\right)\)

\(=x\left(x-6\right)+2\left(x-6\right)=\left(x-6\right)\left(x+2\right)\)

helpppppppppppppppppppppppp

\(\left(3x+2\right)^2-\left(x-6\right)^2=\left(3x+2-x+6\right)\left(3x+2+x-6\right)=\left(2x+8\right)\left(4x-4\right)=8\left(x+4\right)\left(x-1\right)\)

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

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

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

\(=8\left(x-1\right)\left(x+4\right)\)

(x-1)*(x+2)*(x+3)*(x+6)

\(=\left(x^2+5x-6\right)\left(x^2+5x+6\right)\)

\(=\left(x^2+5x\right)^2-36\)

\(\left(x^2+1\right)^2-6\left(x^2+1\right)+9\)   

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

\(=\left(x^2+1-3\right)^2\)   

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

Trả lời:

( x² + 1 ) - 6 ( x² + 1 ) + 9

= ( x² + 1 ) - 2.( x² + 1 ).3 + 3²

=  [ ( x² + 1 ) - 3 ) ]²

= ( x² + 1 - 3 )²

= ( x² - 2 )²

\(x^2+x-2\)

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

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

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

chì làm được 1 cách thôi

x²(x²-6)-x²+9

<=>(x⁴-6x²+9)-x²

<=>(x²-3)²-x²

<=>(x²-3-x)(x²-3+x)

\(x^2-6x+8\)

\(=\left(x^2-6x+9\right)-1\)

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

\(=\left(x-3-1\right)\left(x-3+1\right)\)

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

(Tíck cho mìk vs nha!)

cách 2:

x²  -6x +8 = x² -2x -4x+8= x(x-2) -4(x-2)

= (x-2)(x-4)