\(5\cdot\left(4x-5\right)=4x-5\)
\(\Leftrightarrow20x-25-4x+5=0\)
\(\Leftrightarrow16x-20=0\)
\(\Leftrightarrow x=\dfrac{20}{16}=\dfrac{5}{4}\)
Cách 2:
Ta có: 5x(4x-5)=4x-5
\(\Leftrightarrow5x\left(4x-5\right)-\left(4x-5\right)=0\)
\(\Leftrightarrow\left(4x-5\right)\left(5x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}4x-5=0\\5x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}4x=5\\5x=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{4}\\x=\dfrac{1}{5}\end{matrix}\right.\)
Vậy: \(S=\left\{\dfrac{5}{4};\dfrac{1}{5}\right\}\)
