\(x^2-5x+4=0\)
\(\Leftrightarrow x^2-x-4x+4=0\)
\(\Leftrightarrow x^2-x-\left(4x-4\right)=0\)
\(\Leftrightarrow x\left(x-1\right)-4\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-4\right)\left(x-1\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-4=0\\x-1=0\end{cases}\Rightarrow\orbr{\begin{cases}x=4\\x=1\end{cases}}}\)
Vậy x = { 1; 4 }