1: Thay m=2 vào hệ, ta được:
\(\left\{{}\begin{matrix}2x-y=2\\5x+2y=14\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}4x-2y=4\\5x+2y=14\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}9x=18\\2x-y=2\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=2\\y=2x-2=2\cdot2-2=2\end{matrix}\right.\)
2: Để hệ có nghiệm duy nhất thì \(\dfrac{m}{5}\ne\dfrac{-1}{2}\)
=>\(m\ne-\dfrac{5}{2}\)
\(\left\{{}\begin{matrix}mx-y=2\\5x+2y=14\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}2mx-2y=4\\5x+2y=14\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\left(2m+5\right)=18\\mx-y=2\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=\dfrac{18}{2m+5}\\y=mx-2=\dfrac{18m}{2m+5}-2=\dfrac{18m-4m-10}{2m+5}=\dfrac{14m-10}{2m+5}\end{matrix}\right.\)
\(x+2y=2\)
=>\(\dfrac{18+28m-20}{2m+5}=2\)
=>\(28m-2=4m+10\)
=>24m=12
=>\(m=\dfrac{1}{2}\left(nhận\right)\)
