`2x(x - 5) - x + 5 = 0`
`=> 2x(x-5) - (x - 5) = 0`
`=> (x-5)(2x - 1) = 0`
TH1:
`x - 5 = 0`
`=> x = 5`
TH2:
`2x - 1 = 0`
`=> 2x = 1`
`=> x = 1/2`
Vậy `x = 5` ; `x = 1/2`
\(2x.\left(x-5\right)-x+5=0\)
\(\Leftrightarrow2x\left(x-5\right)-\left(x-5\right)=0\)
\(\Leftrightarrow\left(x-5\right)\left(2x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-5=0\\2x-1=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=\dfrac{1}{2}\end{matrix}\right.\)