ĐKXĐ : \(\left\{{}\begin{matrix}x+1\ne0\\x-2\ne0\end{matrix}\right.\) => \(\left\{{}\begin{matrix}x\ne-1\\x\ne2\end{matrix}\right.\)
Ta có : \(\frac{x+2}{x+1}+\frac{3}{x-2}=\frac{3}{x^2-x-2}+1\)
=> \(\frac{\left(x+2\right)\left(x-2\right)}{\left(x+1\right)\left(x-2\right)}+\frac{3\left(x+1\right)}{\left(x+1\right)\left(x-2\right)}=\frac{3}{\left(x+1\right)\left(x-2\right)}+\frac{\left(x+1\right)\left(x-2\right)}{\left(x+1\right)\left(x-2\right)}\)
=> \(\left(x+2\right)\left(x-2\right)+3\left(x+1\right)=3+\left(x+1\right)\left(x-2\right)\)
=> \(x^2-4+3x+3-3-x^2+x+2=0\)
=> \(4x-2=0\)
=> \(x=\frac{1}{2}\left(TM\right)\)
Vậy phương trình trên có nghiệm là \(S=\left\{\frac{1}{2}\right\}\)
ĐKXĐ: x≠2; x≠-1
Ta có: \(\frac{x+2}{x+1}+\frac{3}{x-2}=\frac{3}{x^2-x-2}+1\)
\(\Leftrightarrow\frac{\left(x+2\right)\left(x-2\right)}{\left(x+1\right)\left(x-2\right)}+\frac{3\left(x+1\right)}{\left(x-2\right)\left(x+1\right)}=\frac{3}{\left(x-2\right)\left(x+1\right)}+\frac{\left(x-2\right)\left(x+1\right)}{\left(x-2\right)\left(x+1\right)}\)
\(\Leftrightarrow x^2-4+3x+3=3+x^2-x-2\)
\(\Leftrightarrow x^2+3x-1-x^2+x-1=0\)
\(\Leftrightarrow4x-2=0\)
\(\Leftrightarrow2\left(2x-1\right)=0\)
Vì 2≠0
nên 2x-1=0
hay \(x=\frac{1}{2}\)(tm)
Vậy: \(x=\frac{1}{2}\)