\(ĐKXĐ:\left\{{}\begin{matrix}x+1\ne0\\x-2\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ne-1\\x\ne2\end{matrix}\right.\)
\(a,\dfrac{x+2}{x+1}+\dfrac{3}{x-2}=\dfrac{3}{x^2-x-2}+1\)
\(\Leftrightarrow\dfrac{\left(x+2\right)\left(x-2\right)+3\left(x+1\right)}{\left(x+1\right)\left(x-2\right)}=\dfrac{3+\left(x+1\right)\left(x-2\right)}{\left(x+1\right)\left(x-2\right)}\)
\(\Leftrightarrow\dfrac{x^2-4+3x+3}{\left(x+1\right)\left(x-2\right)}=\dfrac{3+x^2-x-2}{\left(x+1\right)\left(x-2\right)}\)
\(\Rightarrow x^2+3x-1=x^2-x+1\)
\(\Leftrightarrow x^2+3x-x^2+x=1+1\)
\(\Leftrightarrow4x=2\)
\(\Leftrightarrow x=\dfrac{1}{2}\) ( t/m )
Vậy pt có nghiệm duy nhất \(x=\dfrac{1}{2}\)
