a/4x(x−2)−5(x−2)=0
<=>(4x-5)(x-2)=0<=>4x-5=0->x=5/4 x-2=0->x=2\(4x\left(x-2\right)-5\left(x-2\right)=\left(4x-5\right)\left(x-2\right)=0\)
=> \(\left[{}\begin{matrix}4x-5=0\\x-2=0\end{matrix}\right.\left[{}\begin{matrix}x=\dfrac{5}{4}\\x=2\end{matrix}\right.\)
Ta có: \(4x\left(x-2\right)-5\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(4x-5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{5}{4}\end{matrix}\right.\)