Đáp án:
$\rm a) \ \ x(x-2)=3x-4\\\Leftrightarrow x^2-2x-3x+4=0\\\Leftrightarrow x^2-5x+4=0\\\Leftrightarrow (x-1)(x-4)=0\\\Leftrightarrow \left[ \begin{array}{l}x-1=0\\x-4=0\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}x=1\\x=4\end{array} \right. \\ b) \ \begin{cases}\rm 5x-7=3y\\\rm 2x+y=5\end{cases} \Leftrightarrow \begin{cases}\rm 5x-3y=7\\\rm 6x+3y=15\end{cases} \\ \Leftrightarrow \begin{cases}\rm 11x=22\\\rm y=5-2x\end{cases} \Leftrightarrow \begin{cases}\rm x=2\\\rm y=1\end{cases} \\ Vậy \ (x;y)=(2;1)$
