a)
ĐKXĐ: \(x\notin\left\{1;-1\right\}\)
Ta có: \(\dfrac{5\left(1-x\right)}{x+1}=\dfrac{2x+1}{1-x}\)
\(\Leftrightarrow\dfrac{-5\left(x-1\right)}{x+1}=\dfrac{-2x-1}{x-1}\)
\(\Leftrightarrow-5\left(x-1\right)^2=\left(-2x-1\right)\left(x+1\right)\)
\(\Leftrightarrow-5\left(x^2-2x+1\right)=\left(-2x^2-2x-x-1\right)\)
\(\Leftrightarrow-5x^2+10x-5+2x^2+3x+1=0\)
\(\Leftrightarrow-3x^2+13x-4=0\)
\(\Leftrightarrow-3x^2+12x+x-4=0\)
\(\Leftrightarrow-3x\left(x-4\right)+\left(x-4\right)=0\)
\(\Leftrightarrow\left(x-4\right)\left(-3x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=4\\x=\dfrac{1}{3}\end{matrix}\right.\)
Vậy: \(S=\left\{4;\dfrac{1}{3}\right\}\)
b)
ĐKXĐ: \(x\notin\left\{4;2\right\}\)
Ta có: \(\dfrac{2-x}{4-x}+1=\dfrac{3-x}{x-2}\)
\(\Leftrightarrow\dfrac{x-2+x-4}{x-4}=\dfrac{3-x}{x-2}\)
\(\Leftrightarrow\left(2x-6\right)\left(x-2\right)=\left(3-x\right)\left(x-4\right)\)
\(\Leftrightarrow\left(2x-6\right)\left(x-2\right)+\left(x-3\right)\left(x-4\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(2x-4+x-4\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(3x-8\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=\dfrac{8}{3}\end{matrix}\right.\)
Vậy: \(S=\left\{3;\dfrac{8}{3}\right\}\)
