ĐKXĐ:\(x\ne-2,x\ne3\)
\(\dfrac{2x-3}{x+2}+\dfrac{x-5}{3-x}-\dfrac{10}{x^2-x-6}=2\\ \Leftrightarrow\dfrac{\left(2x-3\right)\left(x-3\right)}{\left(x+2\right)\left(x-3\right)}-\dfrac{\left(x-5\right)\left(x+2\right)}{\left(x+2\right)\left(x-3\right)}-\dfrac{10}{\left(x+2\right)\left(x-3\right)}=\dfrac{2\left(x+2\right)\left(x-3\right)}{\left(x+2\right)\left(x-3\right)}\\ \Leftrightarrow\dfrac{2x^2-9x+9}{\left(x+2\right)\left(x-3\right)}-\dfrac{x^2-3x-10}{\left(x+2\right)\left(x-3\right)}-\dfrac{10}{\left(x+2\right)\left(x-3\right)}-\dfrac{2\left(x^2-x-6\right)}{\left(x+2\right)\left(x-3\right)}=0\)
\(\Leftrightarrow\dfrac{2x^2-9x+9-x^2+3x+10-10-2x^2+2x+12}{\left(x+2\right)\left(x-3\right)}=0\)
\(\Rightarrow-x^2-4x+21=0\\ \Leftrightarrow x^2+4x-21=0\\ \Leftrightarrow\left(x^2-3x\right)+\left(7x-21\right)=0\\ \Leftrightarrow x\left(x-3\right)+7\left(x-3\right)=0\\ \Leftrightarrow\left(x-3\right)\left(x+7\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=3\left(ktm\right)\\x=-7\left(tm\right)\end{matrix}\right.\)
