Để hệ có nghiệm duy nhất thì \(\dfrac{2m}{8}\ne\dfrac{1}{m}\)
=>\(m^2\ne4\)
=>\(m\notin\left\{2;-2\right\}\)
\(\left\{{}\begin{matrix}2mx+y=2\\8x+my=m+2\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}2m^2\cdot x+my=2m\\8x+my=m+2\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x\left(2m^2-8\right)=m-2\\y=2-2mx\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=\dfrac{m-2}{2m^2-8}=\dfrac{1}{2\left(m+2\right)}\\y=2-\dfrac{2m}{2\left(m+2\right)}=2-\dfrac{m}{m+2}=\dfrac{2m+4-m}{m+2}=\dfrac{m+4}{m+2}\end{matrix}\right.\)
4x+3y=7
=>\(\dfrac{4}{2\left(m+2\right)}+\dfrac{3\left(m+4\right)}{m+2}=7\)
=>\(\dfrac{2+3\left(m+4\right)}{m+2}=7\)
=>7(m+2)=2+3m+12
=>7m+14=3m+14
=>4m=0
=>m=0(nhận)
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