ĐKXĐ bạn tự xét nha
\(\dfrac{x+2}{x+1}+\dfrac{3}{x-2}=\dfrac{3}{x^2-x-2}+1\)
\(\dfrac{x+2}{x+1}+\dfrac{3}{x-2}-\dfrac{3}{x^2-x-2}-1=0\)
\(\dfrac{\left(x+2\right)\left(x-2\right)}{\left(x+1\right)\left(x-2\right)}+\dfrac{3\left(x+1\right)}{\left(x+1\right)\left(x-2\right)}-\dfrac{3}{\left(x+1\right)\left(x-2\right)}-\dfrac{\left(x+1\right)\left(x-2\right)}{\left(x+1\right)\left(x-2\right)}=0\)
\(\dfrac{\left(x+2\right)\left(x-2\right)+3\left(x+1\right)-3-\left(x+1\right)\left(x-2\right)}{\left(x+1\right)\left(x-2\right)}=0\)
\(\dfrac{x^2-4+3x+3-3-x^2+x+2}{\left(x+1\right)\left(x-2\right)}=0\)
\(\dfrac{4x-2}{\left(x+1\right)\left(x-2\right)=0}\)
\(\Rightarrow4x-2=0\)
\(\Leftrightarrow x=\dfrac{1}{2}\left(thoa-man\right)\)
Vậy \(x=\dfrac{1}{2}\)
