\(\frac{x+2}{x-2}\)-\(\frac{1}{x}\)=\(\frac{2}{x^2-2x}\)
\(\frac{x+2}{x-2}\)-\(\frac{1}{x}\)=\(\frac{2}{x\left(x-2\right)}\) \(\frac{\left(x+2\right)x}{\left(x-2\right)x}\)-\(\frac{x-2}{x\left(x-2\right)}\)=\(\frac{2}{x\left(x-2\right)}\) x(x+2)-x+2=2 x2+2x-x+2=2 x2+2x-x=2-2 x2+x=0 x(x+1)=0 x=0 hoặc x+1=0 x=0 hoặc x=-1
\(\frac{x+2}{x-2}-\frac{1}{x}=\frac{2}{x^2-2x}\left(x\ne0;x\ne2\right)\)
\(\Leftrightarrow\frac{x\left(x+2\right)}{x\left(x-2\right)}-\frac{x-2}{x\left(x-2\right)}=\frac{2}{x\left(x-2\right)}\)
\(\Rightarrow x^2+2x-x+2-2=0\)
\(\Leftrightarrow x^2+x=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\left(kot/m\right)\\x=-1\left(t/m\right)\end{matrix}\right.\)
Vậy pt có nghiệm x =-1
\(\frac{x+2}{x-2}-\frac{1}{x}=\frac{2}{x^2-2x}\\ \frac{x+2}{x-2}-\frac{1}{x}-\frac{2}{x^2-2x}=0\\ \frac{x\left(x+2\right)}{x\left(x-2\right)}-\frac{x-2}{x\left(x-2\right)}-\frac{2}{x\left(x-2\right)}=0\\ x^2+2x-x+2-2=0\\ x\left(x+1\right)=0\\ x=0,x=-1\)