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:

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

a, \(\frac{5}{2}x^2y^2+15x^2y-30xy^2=5xy\left(\frac{1}{2}xy+3x-6y\right)\)

b, \(16x^2+24x-8xy-6y+y^2\)

\(=\left(16x^2-8xy+y^2\right)+\left(24x-6y\right)=\left(4x-y\right)^2+6\left(4x-y\right)\)

\(=\left(4x-y\right)\left[\left(4x-y\right)+6\right]=\left(4x-y\right)\left(4x-y+6\right)\)

c, \(2x^2-5x-7=2x^2-7x+2x-7\)

\(=2x\left(x+1\right)-7\left(x+1\right)=\left(2x-7\right)\left(x+1\right)\)