Câu 3:
1: x^2-5x+4=0
=>\(\left(x-1\right)\left(x-4\right)=0\)
=>\(\left[{}\begin{matrix}x=1\\x=4\end{matrix}\right.\)
2: \(\left\{{}\begin{matrix}x-2y=1\\2x+y=7\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x-2y=1\\4x+2y=14\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}5x=15\\2x+y=7\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=3\\y=7-2x=1\end{matrix}\right.\)