\(\frac{x-1}{x+2}-\frac{x}{x-2}=\frac{5x+2}{4-x^2}\)
\(\Leftrightarrow\frac{x-1}{x+2}+\frac{x}{2-x}=\frac{5x-2}{4-x^2}\)
ĐKXĐ : \(x\ne\pm2\)
Suy ra \(\left(x-1\right)\left(2-x\right)+\left(x+2\right)x=5x-2\)
\(\Leftrightarrow2x-x^2-2+x+x^2+2x=5x-2\)
\(\Leftrightarrow5x-2=5x-2\)
\(\Leftrightarrow0x=0\)
Vậy phương trình vô số nghiệm
\(ĐK:x\ne\pm2\)
\(\frac{x-1}{x+2}-\frac{x}{x-2}=\frac{x^2-3x+2}{x^2-4}-\frac{x^2-2x}{x^2-4}=\frac{-2x^2-x+2}{x^2-4}=\frac{2x^2+x-2}{4-x^2}=\frac{5x-2}{4-x^2}\Leftrightarrow2x^2+x-2=5x-2\Leftrightarrow2x^2+x=5x\Leftrightarrow2x^2-4x=0\Leftrightarrow x^2-2x=0\Leftrightarrow x\left(x-2\right)=0\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\left(loai\right)\\x=0\end{matrix}\right.\)
\(\frac{x-1}{x+2}-\frac{x}{x-2}=\frac{5x-2}{4-x^2}\)
ĐK: x\(\ne\)2; x\(\ne\)-2
\(\Leftrightarrow\frac{x-1}{x+2}-\frac{x}{x-2}=-\frac{5x-2}{x^2-4}\)
\(\Leftrightarrow\frac{\left(x-1\right)\left(x-2\right)}{x^2-4}-\frac{x\left(x+2\right)}{x^2-4}=-\frac{5x-2}{x^2-4}\)
\(\Leftrightarrow\frac{\left(x-1\right)\left(x-2\right)}{x^2-4}-\frac{x\left(x+2\right)}{x^2-4}+\frac{5x-2}{x^2-4}=0\)
\(\Rightarrow x^2-2x-x+2-x^2-2x+5x-2=0\)
\(\Leftrightarrow0x=0\)
Vậy: phương trình trên vô nghiệm