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

Đặt \(x^2-3x-1=a\), ta có:

\(a^2-12a+27=a^2-9a-3a+27=a\left(a-9\right)-3\left(a-9\right)=\left(a-9\right)\left(a-3\right)\)

\(=\left(x^2-3x-1-9\right)\left(x^2-3x-1-3\right)=\left(x^2-3x-10\right)\left(x^2-3x-4\right)\)

Mà \(x^2-3x-10=x^2-5x+2x-10=x\left(x-5\right)+2\left(x-5\right)=\left(x-5\right)\left(x+1\right)\)

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

\(\Rightarrow\left(x^2-3x-1\right)^2-12\left(x^2-3x-1\right)+27=\left(x-5\right)\left(x-4\right)\left(x+1\right)\left(x+2\right)\)

Đặt \(x^2-3x-1=a\)thay vào biểu thức ta được :

\(a^2-12a+27\)

\(=a^2-3a-9a+27\)

\(=a\left(a-3\right)-9\left(a-3\right)\)

\(=\left(a-3\right)\left(a-9\right)\)(1)

Thay \(a=x^2-3x-1\)vào (1) ta được :

\(\left(x^2-3x-1-3\right)\left(x^2-3x-1-10\right)\)

\(=\left(x^2-3x-4\right)\left(x^2-3x-11\right)\)

Bạn Châu sai đáp án cuối

phải là (x²-3x-4)(x²-3x-10) nha

\(9(x-y)^2-27(y-x)^3\\=9(x-y)^2+27(x-y)^3\\=(x-y)^2[9+27(x-y)]\\=(x-y)^2(9+27x-27y)\\=9(x-y)^2(1+3x-3y)\)

= (x³+3x²+3x+1)-(4y)³

=(x+1)³-(4y)³

=(x+1-4y)[(x+1)²+(x+1).4y+16y² ]

=(x+1-4y)[(x²+2x+1)+(4xy+4y)+16y²]

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

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

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

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

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

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

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

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

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

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

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

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

hk tốt

x³+27+(x+3)(x+9)

= (x+3)(x²-3x+9)+(x+3)(x+9)

= (x+3)(x²-3x+9+x+9)

=(x+3)(x²-2x+18)

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

x³+27+(x+3)(x+9)

= (x+3)(x²-3x+9)+(x+3)(x+9)

= (x+3)(x²-3x+9+x+9)

=(x+3)(x²-2x+18)

\(-x^3+9x^2-27x+27=\left(3-x\right)^3\)

\(-x^3+9x^2-27x+27\)

\(=-x^3+3x^2+6x^2-18x-9x+27\)

\(=-x^2\left(x-3\right)+6x\left(x-3\right)-9\left(x-3\right)\)

\(=-\left(x-3\right)\left(x^2-6x+9\right)\)

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

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