a) Thay m = 2 , ta có :
\(\left\{{}\begin{matrix}-2.2x+y=4\\2x+2y=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-4x+y=4\\4x+4y=4\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}5y=8\\4x+4y=4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=\frac{8}{5}\\4x+\frac{4.8}{5}=4\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y=\frac{8}{5}\\4x=-\frac{12}{5}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=\frac{8}{5}\\x=-\frac{3}{5}\end{matrix}\right.\)
b) Tính x,y theo m :
\(\left\{{}\begin{matrix}-2mx+y=4\\2x+my=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-2mx+y=4\\2mx+m^2y=2m\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(m^2+1\right)y=4+2m\\2x+my=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=\frac{4+2m}{m^2+1}\\2x+\frac{4m+2m^2}{m^2+1}=2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y=\frac{4+2m}{m^2+1}\\2x=\frac{2m^2+2-4m-2m^2}{m^2+1}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=\frac{4+2m}{m^2+1}\\x=\frac{1-2m}{m^2+1}\end{matrix}\right.\)
\(\Rightarrow S=x+y=\frac{1-2m}{m^2+1}+\frac{4+2m}{m^2+1}=\frac{5}{m^2+1}\)
Để S đạt GTLN thì :
m2 + 1 có GTNN mà m2 + 1 ≥ 1
⇒ S ≤ 5 . Dấu = xảy ra khi m2 + 1 = 1 ⇔ m = 0
Vậy . . . . . . . . .
