\(\frac{x+2}{x+1}+\frac{3}{x-2}=\frac{3}{x^2-x-x}+1\)
\(\Leftrightarrow\frac{\left(x+2\right)\left(x-2\right)}{\left(x+1\right)\left(x-2\right)}+\frac{3\left(x+1\right)}{\left(x-2\right)\left(x+1\right)}=\frac{3}{\left(x+1\right)\left(x-2\right)}+\frac{\left(x+1\right)\left(x-2\right)}{\left(x+1\right)\left(x-2\right)}\)
\(\Rightarrow x^2-4+3x+3=3+x^2-2x+x-2\)
\(\Leftrightarrow x^2-x^2+3x+2x-x=1+4-3\)
\(\Leftrightarrow4x=2\)
\(\Leftrightarrow x=\frac{1}{2}\)
\(ĐKXĐ:\hept{\begin{cases}x+1\ne0\\x-2\ne0\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x\ne-1\\x\ne2\end{cases}}\)
Vậy nên x=1/2 thỏa mãn ĐKXĐ nhé!