\(a,\dfrac{3x-2}{4}+\dfrac{x+3}{2}=\dfrac{x-1}{3}-\dfrac{-x-1}{12}\)
\(\Leftrightarrow3\left(3x-2\right)+6\left(x+3\right)=4\left(x-1\right)+x+1\)
\(\Leftrightarrow9x-6+6x+18=4x-4+x+1\)
\(\Leftrightarrow9x+6x-4x-x=1+6-18-4\)
\(\Leftrightarrow10x=-15\)
\(\Leftrightarrow x=-\dfrac{3}{2}\)
Vậy \(S=\left\{-\dfrac{3}{2}\right\}\)
\(b,\left(3x+1\right)\left(x-2\right)=\left(x-2\right)\left(x+1\right)\)
\(\Leftrightarrow\left(3x+1\right)\left(x-2\right)-\left(x-2\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(3x+1-x-1\right)=0\)
\(\Leftrightarrow2x\left(x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\end{matrix}\right.\)
Vậy \(S=\left\{0;2\right\}\)
\(c,\dfrac{2}{x-1}+\dfrac{2}{x+1}-\dfrac{2x^2+2}{x^2-1}=0\left(ĐKXĐ:x\ne1;x\ne-1\right)\)
\(\Leftrightarrow2\left(x+1\right)+2\left(x-1\right)-2x^2+2=0\)
\(\Leftrightarrow2x+2+2x-2-2x^2-2=0\)
\(\Leftrightarrow-2x^2+4x-2=0\)
\(\Leftrightarrow-2x^2+2x+2x-2=0\)
\(\Leftrightarrow-2x\left(x-2\right)+2\left(x-2\right)=0\)
\(\Leftrightarrow\left(-2x+2\right)\left(x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}-2x+2=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\left(loại\right)\\x=2\left(nhận\right)\end{matrix}\right.\)
Vậy \(S=\left\{2\right\}\)