Để hệ có nghiệm duy nhất thì \(\dfrac{m}{3}\ne\dfrac{-1}{m}\)
=>\(m^2\ne-3\)(luôn đúng)
=>Hệ luôn có nghiệm duy nhất
\(\left\{{}\begin{matrix}mx-y=2\\3x+my=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=mx-2\\3x+m\left(mx-2\right)=5\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}y=mx-2\\3x+m^2x-2m=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\left(m^2+3\right)=2m+5\\y=mx-2\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=\dfrac{2m+5}{m^2+3}\\y=m\cdot\dfrac{2m+5}{m^2+3}-2=\dfrac{2m^2+5m-2m^2-6}{m^2+3}=\dfrac{5m-6}{m^2+3}\end{matrix}\right.\)
\(x+y=\dfrac{3}{m^2+3}\)
=>\(\dfrac{2m+5+5m-6}{m^2+3}=\dfrac{3}{m^2+3}\)
=>7m-1=3
=>7m=4
=>\(m=\dfrac{4}{7}\)
=>\(\dfrac{a}{b}=\dfrac{4}{7}\)
=>a=4;b=7
=>a+b=11
