\(\dfrac{1}{x-2}+\dfrac{1}{x^2+x-6}=\dfrac{1}{x+3}-1\left(đkxđ:x\ne2;x\ne-3\right)\\ \Leftrightarrow\dfrac{1}{x-2}+\dfrac{1}{\left(x-2\right)\left(x+3\right)}=\dfrac{1}{x+3}-1\\ \Leftrightarrow\dfrac{x+3}{\left(x-2\right)\left(x+3\right)}+\dfrac{1}{\left(x-2\right)\left(x+3\right)}=\dfrac{x-2}{\left(x-2\right)\left(x+3\right)}-\dfrac{\left(x-2\right)\left(x+3\right)}{\left(x-2\right)\left(x+3\right)}\\ \Rightarrow x+3+1=x-2-\left(x^2+3x-2x-6\right)\\ \Leftrightarrow x+4=x-2-x^2-3x+2x+6\\ \Leftrightarrow x-x+x^2+3x-2x=-2+6-4\\ \Leftrightarrow x^2+x=0\)
\(\Leftrightarrow x\left(x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x+1=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=-1\end{matrix}\right.\)
