\(\left\{{}\begin{matrix}\left(m+1\right)x-y=3\\mx+y=m\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\left(2m+1\right)x=m+3\\mx+y=m\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=\frac{m+3}{2m+1}\\\frac{m\left(m+3\right)}{2m+1}+y=m\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\frac{m+3}{2m+1}\\y=\frac{m^2-2m}{2m+1}\end{matrix}\right.\)
a) Thay \(m=\sqrt{2}\) ta có :
\(\left\{{}\begin{matrix}x=\frac{\sqrt{2}+3}{2\sqrt{2}+1}\\y=\frac{2-2\sqrt{2}}{2\sqrt{2}+1}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\frac{1+5\sqrt{2}}{7}\\y=\frac{6\sqrt{2}-10}{7}\end{matrix}\right.\)
b) Để x + y > 0 thì :
\(\frac{m+3}{2m+1}+\frac{m^2-2m}{2m+1}>0\Leftrightarrow\frac{m^2-m+3}{2m+1}>0\)
Lại có :
\(m^2-m+3=m^2-m+\frac{1}{4}+\frac{11}{4}=\left(m-\frac{1}{2}\right)^2+\frac{11}{4}>0\)
\(\Rightarrow2m+1>0\Leftrightarrow m>-\frac{1}{2}\)
Thay x = y = a vào hệ , ta có :
\(a=\frac{m+3}{2m+1}=\frac{m^2-2m}{2m+1}\Rightarrow m+3=m^2-2m\)
\(\Leftrightarrow m^2-3m-3=0\left(1\right)\)
Δ = 9 + 4.3.1 = 21 > 0
Vậy pt có 2 nghiệm phân biệt :
\(m_1=\frac{3+\sqrt{21}}{2}\left(tm\right);m_2=\frac{3-\sqrt{21}}{2}\left(tm\right)\)
Để hệ có nghiệm duy nhất thỏa mãn x + y > 0 thì m = ...
