\(\Leftrightarrow\left(x-1\right)\left(x-2\right)+6x=-\left(x+1\right)\left(x+2\right)\)
\(\Leftrightarrow x^2-3x+2+6x=-x^2-3x-2\)
\(\Leftrightarrow2x^2+6x+4=0\)
=>(x+1)(x+2)=0
=>x=-1
\(\dfrac{x-1}{x+2}+\dfrac{6x}{x^2-4}=\dfrac{x+1}{2-x}\);\(ĐK:x\ne\pm2\)
\(\Leftrightarrow\dfrac{\left(x-1\right)\left(x-2\right)+6x}{\left(x-2\right)\left(x+2\right)}=-\dfrac{\left(x+1\right)\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}\)
\(\Leftrightarrow\left(x-1\right)\left(x-2\right)+6x=-\left(x+1\right)\left(x+2\right)\)
\(\Leftrightarrow x^2-2x-x+2+6x=-\left(x^2+2x+x+2\right)\)
\(\Leftrightarrow2x^2+6x+4=0\)
\(\Leftrightarrow2x^2+2x+4x+4=0\)
\(\Leftrightarrow2x\left(x+1\right)+4\left(x+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(2x+4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-1\left(tm\right)\\x=-2\left(ktm\right)\end{matrix}\right.\)
Vậy \(S=\left\{-1\right\}\)
