Vì \(\dfrac{2}{3}\ne\dfrac{-1}{2}\)
nên hệ luôn có nghiệm duy nhất
\(\left\{{}\begin{matrix}2x+y=m\\3x-2y=5\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}4x+2y=2m\\3x-2y=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}4x+2y+3x-2y=2m+5\\2x+y=m\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}7x=2m+5\\y=m-2x\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=\dfrac{2}{7}m+\dfrac{5}{7}\\y=m-2\left(\dfrac{2}{7}m+\dfrac{5}{7}\right)=\dfrac{3}{7}m-\dfrac{10}{7}\end{matrix}\right.\)
Vậy: \(M\left(\dfrac{2}{7}m+\dfrac{5}{7};\dfrac{3}{7}m-\dfrac{10}{7}\right)\)
Để M nằm hoàn toàn phía bên trái đường thẳng \(x=\sqrt{3}\) thì \(\dfrac{2}{7}m+\dfrac{5}{7}< \sqrt{3}\)
=>\(2m+5< 3\sqrt{7}\)
=>\(2m< 3\sqrt{7}-5\)
=>\(m< \dfrac{3\sqrt{7}-5}{2}\)
