\(\frac{-x-x+12}{x^2-6x+9}\cdot\frac{2x-6}{x+4}\)

\(=\frac{-2x+12}{\left(x-3\right)^2}\cdot\frac{2\left(x-3\right)}{x+4}\)

\(=\frac{-2\left(x-6\right)}{\left(x-3\right)^2}\cdot\frac{2\left(x-3\right)}{x+4}=\frac{-2\left(x-6\right)\cdot2\left(x-3\right)}{\left(x-3\right)^2\left(x+4\right)}=\frac{-4\left(x-6\right)\left(x-3\right)}{\left(x-3\right)^2\left(x+4\right)}=\frac{-4\left(x-6\right)}{\left(x-3\right)\left(x+4\right)}\)

\(A=\frac{x+6}{x-2}\)ĐKXĐ : \(x\ne2\)

\(=\frac{x-2+8}{x-2}=\frac{8}{x-2}\)

Suy ra : \(x-2\inƯ\left(8\right)=\left\{\pm1;\pm2;\pm4;\pm8\right\}\)

x+x=1

vậy x= 1+0 hoặc 3 +(-2)

ok kết bạn với mình nha

=)) Đấy là cái đề Tìm x á 

\(10x\left(x-5\right)-x+5=0\)

\(\Leftrightarrow10x\left(x-5\right)-\left(x-5\right)=0\)

\(\Leftrightarrow\left(10x-1\right)\left(x-5\right)=0\Leftrightarrow x=\frac{1}{10};5\)

bn ấy ko cách ra lm mk hiểu nhầm

a, \(\frac{x}{xy-y^2}+\frac{2x-y}{xy-x^2}=\frac{x}{y\left(x-y\right)}+\frac{2x-y}{x\left(y-x\right)}\)

\(=\frac{x^2}{xy\left(x-y\right)}-\frac{2xy-y^2}{xy\left(x-y\right)}=\frac{\left(x-y\right)^2}{xy\left(x-y\right)}=\frac{x-y}{xy}\)

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

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