a: Để hệ có nghiệm duy nhất thì \(\frac{m}{2}<>\frac{-2}{-m}\)
=>\(m^2<>4\)
=>m∉{2;-2}
\(\begin{cases}mx-2y=2m-1\\ 2x-my=9-3m\end{cases}\Rightarrow\begin{cases}2mx-4y=4m-2\\ 2mx-m^2y=9m-3m^2\end{cases}\)
=>\(\begin{cases}2mx-4y-2mx+m^2y=4m-2-9m+3m^2\\ 2x-my=9-3m\end{cases}\)
=>\(\begin{cases}y\left(m^2-4\right)=3m^2-5m-2=3m^2-6m+m-2=\left(m-2\right)\left(3m+1\right)\\ 2x-my=9-3m\end{cases}\)
=>\(\begin{cases}y=\frac{\left(m-2\right)\left(3m+1\right)}{\left(m-2\right)\left(m+2\right)}=\frac{3m+1}{m+2}\\ 2x=my+9-3m=\frac{m\left(3m+1\right)+\left(9-3m\right)\left(m+2\right)}{m+2}=\frac{3m^2+m+9m+18-3m^2-6m}{m+2}=\frac{4m+18}{m+2}\end{cases}\)
=>\(\begin{cases}y=\frac{3m+1}{m+2}\\ x=\frac{2m+9}{m+2}\end{cases}\)
b: \(S=x^2+y^2\)
\(=\frac{9m^2+6m+1+4m^2+36m+81}{\left(m+2\right)^2}=\frac{13m^2+39m+82}{m^2+4m+4}\)
\(=\frac{13m^2+52m+52-13m+30}{m^2+4m+4}=13+\frac{-13m-26+56}{\left(m+2\right)^2}\)
\(=13-\frac{13}{m+2}+\frac{56}{\left(m+2\right)^2}\)
\(=56\left\lbrack\frac{1}{\left(m+2\right)^2}-\frac{13}{56}\cdot\frac{1}{m+2}+\frac{13}{56}\right\rbrack\)
\(=56\cdot\left\lbrack\frac{1}{\left(m+2\right)^2}-2\cdot\frac{1}{m+2}\cdot\frac{13}{112}+\frac{169}{12544}+\frac{2743}{12544}\right\rbrack=56\left(\frac{1}{m+2}-\frac{13}{112}\right)^2+\frac{2743}{224}\ge\frac{2743}{224}\) với mọi m
Dấu '=' xảy ra khi \(\frac{1}{m+2}-\frac{13}{112}=0\)
=>\(m+2=\frac{112}{13}\)
=>\(m=\frac{112}{13}-2=\frac{86}{13}\) (nhận)
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