\(\dfrac{1}{x^2-3x+2}-\dfrac{1}{x-2}=2\left(đkxđ:x\ne1;x\ne2\right)\\ \Leftrightarrow\dfrac{1}{\left(x-2\right)\left(x-1\right)}-\dfrac{1}{x-2}=2\\ \Leftrightarrow\dfrac{1}{\left(x-2\right)\left(x-1\right)}-\dfrac{x-1}{\left(x-2\right)\left(x-1\right)}=\dfrac{2\left(x-2\right)\left(x-1\right)}{\left(x-2\right)\left(x-1\right)}\\ \Rightarrow1-x+1=\left(2x-4\right)\left(x-1\right)\\ \Leftrightarrow2-x=2x^2-2x-4x+4\\ \Leftrightarrow-x-2x^2+2x+4x=4-2\\ \Leftrightarrow-2x^2+5x-2=0\\ \Leftrightarrow-\left(2x^2-x-4x+2\right)=0\\ \Leftrightarrow-\left[\left(2x^2-x\right)-\left(4x-2\right)\right]=0\)
\(\Leftrightarrow-\left[x\left(2x-1\right)-2\left(2x-1\right)\right]=0\\ \Leftrightarrow-\left(2x-1\right)\left(x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}-2x+1=0\\x-2=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}-2x=-1\\x=2\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=2\end{matrix}\right.\)