\(4x^2-1=\left(x-5\right)\left(1-2x\right)\)
\(\Leftrightarrow4x^2-1=x-2x^2-5+10x\)
\(\Leftrightarrow6x^2-11x+4=0\)
\(\Leftrightarrow6x^2-8x-3x+4=0\)
\(\Leftrightarrow2x\left(3x-4\right)-1\left(3x-4\right)=0\)
\(\Leftrightarrow\left(3x-4\right)\left(2x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{4}{3}\\x=\dfrac{1}{2}\end{matrix}\right.\)
\(4x^2-1=\left(x-5\right)\left(1-2x\right)\)
\(\Leftrightarrow\left(2x-1\right)\left(2x+1\right)+\left(x-5\right)\left(2x-1\right)=0\)
\(\Leftrightarrow\left(2x-1\right)\left(2x+1+x-5\right)=0\)
\(\Leftrightarrow\left(2x-1\right)\left(3x-4\right)=0\)
\(\Leftrightarrow2x-1=0\) hay \(3x-4=0\)
\(\Leftrightarrow x=\dfrac{1}{2}\) hay \(x=\dfrac{4}{3}\)
-Vậy \(S=\left\{\dfrac{1}{2};\dfrac{4}{3}\right\}\)
