Sửa đề : \(x^2-xz-9y^2+3yz=\left(x^2-9y^2\right)+\left(-xz+3yz\right)\)

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

\(=\left(x-3y\right)\left[\left(x+3y\right)-z\right]=\left(x-3y\right)\left(x+3y-z\right)\)

\(x^2-xz-9y^2+3yz\)

\(=\left(x^2-9y^2\right)-\left(xz-3yz\right)\)

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

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

\(\left(x^2+10x\right)\left(x^2+10x+24\right)+128\)

\(=x^4+10x^3+24x^2+10x^3+100x^2+240x+128\)

\(=x^4+20x^3+124x^2+240x+128\)

hong biet=))

ta có : 9-9 = 6-6

=> 3(3-3) = 2(3-3)

=> 3=2 

=> 3 = 1+1

giúp mình với plsss

Bài làm 

a, \(A=\frac{x^3-2x^2+x}{x^3-x}=\frac{x\left(x^2-2+1\right)}{x\left(x^2-1\right)}=\frac{x\left(x-1\right)^2}{x\left(x-1\right)\left(x+1\right)}=\frac{x-1}{x+1}\)

ĐK : \(x\ne0;\pm1\)

a,\(ĐKXĐ:x^3-x\ne0\)

            \(\Leftrightarrow x\left(x^2-1\right)\ne0\)

            \(\Leftrightarrow x\left(x-1\right)\left(x+1\right)\ne0\)

           \(\Leftrightarrow\hept{\begin{cases}x\ne0\\x-1\ne0\\x+1\ne0\end{cases}\Leftrightarrow\hept{\begin{cases}x\ne0\\x\ne1\\x\ne-1\end{cases}}}\)

c,\(\frac{x^3-2x^2+x}{x^3-x}=\frac{x\left(x^2-2x+1\right)}{x\left(x^2-1\right)}=\frac{x\left(x-1\right)^2}{x\left(x-1\right)\left(x+1\right)}=\frac{x\left(x-1\right)}{x\left(x+1\right)}\)

Tại \(x=-\frac{1}{3}\)ta có:

\(\frac{-\frac{1}{3}\left(-\frac{1}{3}-1\right)}{-\frac{1}{3}\left(-\frac{1}{3}+1\right)}=-2\)