b: Để hệ có nghiệm duy nhất thì \(\frac{1}{m}<>\frac{-m}{1}\)
=>\(-m^2<>1\)
=>\(m^2<>-1\) (luôn đúng)
=>(I) luôn có nghiệm duy nhất
\(\begin{cases}x-my=2-4m\\ mx+y=3m+1\end{cases}=>\begin{cases}mx-m^2y=2m-4m^2\\ mx+y=3m+1\end{cases}\)
=>\(\begin{cases}mx-m^2y-mx-y=2m-4m^2-3m-1\\ x-my=2-4m\end{cases}\)
=>\(\begin{cases}y\left(-m^2-1\right)=-4m^2-m-1\\ x=my+2-4m\end{cases}\Rightarrow\begin{cases}y=\frac{4m^2+m+1}{m^2+1}\\ x=m\cdot\frac{4m^2+m+1}{m^2+1}+2-4m\end{cases}\)
=>\(\begin{cases}y=\frac{4m^2+m+1}{m^2+1}\\ x=\frac{4m^3+m^2+m+\left(2-4m\right)\left(m^2+1\right)}{m^2+1}=\frac{4m^3+m^2+m+2m^2+2-4m^3-4m}{m^2+1}=\frac{3m^2-3m+2}{m^2+1}\end{cases}\)
=>\(\begin{cases}y=\frac{4m^2+4+m-3}{m^2+1}=4+\frac{m-3}{m^2+1}\\ x=\frac{3m^2+3-3m-1}{m^2+1}=3+\frac{-3m-1}{m^2+1}\end{cases}\)
\(x_0^2+y_0^2-5\left(x_0+y_0\right)\)
\(=\left(4+\frac{m-3}{m^2+1}\right)^2+\left(3+\frac{-3m-1}{m^2+1}\right)^2-5\left(4+\frac{m-3}{m^2+1}+3+\frac{-3m-1}{m^2+1}\right)\)
\(=16+\frac{8\left(m-3\right)}{m^2+1}+\frac{m^2-6m+9}{\left(m^2+1\right)^2}+9-\frac{6\left(3m+1\right)}{m^2+1}+\frac{9m^2+6m+1}{\left(m^2+1\right)^2}\) -5(7+\(\frac{m-3-3m-1}{m^2+1}\) )
\(=25+\frac{8m-24-18m-6}{m^2+1}+\frac{10m^2+10}{m^2+1}-5\left(7+\frac{-2m-4}{m^2+1}\right)\)
\(=35+\frac{-10m-30}{m^2+1}-35+\frac{10m+20}{m^2+1}=\frac{-10m-30+10m+20}{m^2+1}=\frac{-10}{m^2+1}\)
=>\(x_0^2+y_0^2-5\left(x_0+y_0\right)+10=\frac{-10}{m^2+1}+10=\frac{-10+10m^2+10}{m^2+1}=\frac{10m^2}{m^2+1}\)
