a: Thay m=2 vào hệ, ta được:
\(\left\{{}\begin{matrix}2x+y=5\\3x-2y=4\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}4x+2y=10\\3x-2y=4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}7x=14\\2x+y=5\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=2\\y=5-2x=5-2\cdot2=1\end{matrix}\right.\)
b: Để hệ phương trình có nghiệm duy nhất thì \(\dfrac{m}{3}\ne\dfrac{1}{-m}\)
=>\(m^2\ne-3\)(luôn đúng)
\(\left\{{}\begin{matrix}mx+y=5\\3x-my=4\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}y=5-mx\\3x-m\left(5-mx\right)=4\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}y=5-mx\\3x-5m+m^2x=4\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}y=5-mx\\x\left(m^2+3\right)=5m+4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{5m+4}{m^2+3}\\y=5-\dfrac{5m^2+4m}{m^2+3}\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=\dfrac{5m+4}{m^2+3}\\y=\dfrac{5m^2+15-5m^2-4m}{m^2+3}=\dfrac{-4m+15}{m^2+3}\end{matrix}\right.\)
Để x>0 và y>0 thì \(\left\{{}\begin{matrix}\dfrac{5m+4}{m^2+3}>0\\\dfrac{-4m+15}{m^2+3}>0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}5m+4>0\\-4m+15>0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}m>-\dfrac{4}{15}\\m< \dfrac{15}{4}\end{matrix}\right.\)
=>\(-\dfrac{4}{15}< m< \dfrac{15}{4}\)
