\(\dfrac{x^2+2x-3}{x^2-x-6}+\dfrac{x-3}{2-x}=1\) ĐKXĐ:\(x\ne2,x\ne-3\)
\(\Leftrightarrow\dfrac{x^2+2x-3}{x^2-x-2-4}-\dfrac{x-3}{x-2}=1\)
\(\Leftrightarrow\dfrac{x^2+2x-3}{\left(x^2-4\right)+\left(x-2\right)}-\dfrac{x-3}{x-2}=1\)
\(\Leftrightarrow\dfrac{x^2+2x-3}{\left(x-2\right)\left(x+2\right)+\left(x-2\right)}-\dfrac{x-3}{x-2}=1\)
\(\Leftrightarrow\dfrac{x^2+2x-3}{\left(x-2\right)\left(x+3\right)}-\dfrac{x-3}{x-2}=1\)
\(\Leftrightarrow x^2+2x-3-\left(x-3\right)\left(x+3\right)=\left(x-2\right)\left(x+3\right)\)
\(\Leftrightarrow x^2+2x-3-x^2+9=x^2+3x-2x-6\)
\(\Leftrightarrow2x+6=x^2+x-6\)
\(\Leftrightarrow x^2+x-2x-6-6=0\)
\(\Leftrightarrow x^2-x-12=0\)
\(\Leftrightarrow x^2-x-3-9=0\)
\(\Leftrightarrow\left(x^2-9\right)-\left(x+3\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(x+3\right)-\left(x+3\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left(x-4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+3=0\\x-4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\left(loai\right)\\x=4\end{matrix}\right.\)
\(\Rightarrow S=\left\{4\right\}\)