a: Thay m=1 vào hệ, ta được:
\(\left\{{}\begin{matrix}x+y=3\\4x+y=6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-3x=-6\\x+y=3\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=2\\y=3-x=3-2=1\end{matrix}\right.\)
b: Để hệ có nghiệm duy nhất thì \(\dfrac{m}{4}\ne\dfrac{1}{m}\)
=>\(m^2\ne4\)
=>\(m\notin\left\{2;-2\right\}\)
\(\left\{{}\begin{matrix}mx+y=3\\4x+my=6\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}y=-mx+3\\4x+m\left(-mx+3\right)=6\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}y=-mx+3\\x\left(-m^2+4\right)=6-3m\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=\dfrac{3m-6}{m^2-4}=\dfrac{3}{m+2}\\y=\dfrac{-3m}{m+2}+3=\dfrac{-3m+3m+6}{m+2}=\dfrac{6}{m+2}\end{matrix}\right.\)
x>1;y>0
=>\(\left\{{}\begin{matrix}\dfrac{3}{m+2}>1\\\dfrac{6}{m+2}>0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{3-m-2}{m+2}>0\\m+2>0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\dfrac{m-1}{m+2}< 0\\m>-2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-2< m< 1\\-2< m\end{matrix}\right.\)
=>-2<m<1
