a) Khi m=3 thì ta có hệ phương trình:
\(\left\{{}\begin{matrix}3x+2y=4\\2x-y=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}3x+2y=4\\4x-2y=6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}7x=10\\3x+2y=4\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{10}{7}\\3.\dfrac{10}{7}+2y=4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{10}{7}\\y=-\dfrac{1}{7}\end{matrix}\right.\)
Vậy hệ phương trình có nghiệm (x;y)=(\(\dfrac{10}{7}\);\(-\dfrac{1}{7}\))
b) \(\left\{{}\begin{matrix}3x+2y=4\left(1\right)\\2x-y=m\left(2\right)\end{matrix}\right.\)
⇔\(\left\{{}\begin{matrix}3x+2y=4\\y=2x-m\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}3x+2\left(2x-m\right)=4\\y=2x-m\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}7x-2m=4\\y=2x-m\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{2m+4}{7}\\y=\dfrac{4m+8}{7}-m\end{matrix}\right.\)
\(\left\{{}\begin{matrix}x=\dfrac{2m+4}{7}\\y=\dfrac{-3m+8}{7}\end{matrix}\right.\)
mà \(x< 1;y< 1\)nên \(\left\{{}\begin{matrix}\dfrac{2m+4}{7}< 1\\\dfrac{-3m+8}{7}< 1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{2m+4}{7}-1< 0\\\dfrac{-3m+8}{7}-1< 0\end{matrix}\right.\)
⇔\(\left\{{}\begin{matrix}\dfrac{2m-3}{7}< 0\\\dfrac{-3m+1}{7}< 0\end{matrix}\right.\)
mà \(7>0\) nên ta có :
\(\left\{{}\begin{matrix}2m-3< 0\\-3m+1< 0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}m< \dfrac{3}{2}\\m>\dfrac{1}{3}\end{matrix}\right.\)
Vậy \(\dfrac{1}{3}< m< \dfrac{3}{2}\)thì \(x< 0;y< 0\)