a) Thay $m=5$ vào hệ phương trình ta được
\(\left\{{}\begin{matrix}3x+2y=4\\2x-y=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}3x+2y=4\\4x-2y=10\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}7x=14\\y=2x-5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=-1\end{matrix}\right.\)
b) \(\left\{{}\begin{matrix}3x+2y=4\\2x-y=m\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}3x+4x=4+2m\\y=2x-m\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x=\frac{4+2m}{7}\\y=\frac{2\left(4+2m\right)}{7}-m\end{matrix}\right.\)
Theo đề, \(\left\{{}\begin{matrix}x=\frac{4+2m}{7}< 1\\y=\frac{2\left(4+2m\right)}{7}-m< 1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}4+2m< 7\\8+4m-7m< 7\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}m< \frac{3}{2}\\m>1\end{matrix}\right.\)
Vậy với \(1< m< \frac{3}{2}\) thic hệ có nghiệm (x;y) thõa mãn $x<1;y<1$
