Để hệ có nghiệm duy nhất thì \(\dfrac{2m}{5}\ne\dfrac{-5}{-2m}\)
=>\(4m^2\ne25\)
=>\(m^2\ne\dfrac{25}{4}\)
=>\(m\notin\left\{\dfrac{5}{2};-\dfrac{5}{2}\right\}\)
\(\left\{{}\begin{matrix}2mx-5y=-2\\5x-2my=3-2m\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}10mx-25y=-10\\10mx-4m^2y=2m\left(3-2m\right)\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}10mx-25y-10mx+4m^2y=-10-2m\left(3-2m\right)\\2mx-5y=-2\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}y\left(4m^2-25\right)=-10-6m+4m^2=4m^2-6m-10=\left(2m-5\right)\left(m+1\right)\\2mx=5y-2\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}y=\dfrac{m+1}{2m+5}\\2mx=\dfrac{5\left(m+1\right)}{2m+5}-2=\dfrac{5m+5-4m-10}{2m+5}=\dfrac{m-5}{2m+5}\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}y=\dfrac{m+1}{2m+5}\\x=\dfrac{m-5}{2m\left(2m+5\right)}\end{matrix}\right.\)
Để x,y đều nguyên thì \(\left\{{}\begin{matrix}m+1⋮2m+5\\m-5⋮2m\left(2m+5\right)\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2m+2⋮2m+5\\m-5⋮2m\left(2m+5\right)\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}2m+5-3⋮2m+5\\m-5⋮2m\left(2m+5\right)\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-3⋮2m+5\\m-5⋮2m\left(2m+5\right)\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}2m+5\in\left\{1;-1;3;-3\right\}\\m-5⋮2m\left(2m+5\right)\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}m\in\left\{-2;-3;-1;-4\right\}\\m-5⋮2m\left(2m+5\right)\end{matrix}\right.\)
=>m=-1
