\(\left\{{}\begin{matrix}mx+y=7\\2x-y=-4\end{matrix}\right.\left(1\right)\)
Ta có: \(2x-y=-4\)
\(\Rightarrow y=2x+4\)
\(P=x^2+y^2=x^2+\left(2x+4\right)^2=x^2+4x^2+16x+16\)
\(P=5x^2+16x+16=5\left(x^2+2.\frac{8}{5}x+\frac{64}{25}\right)+\frac{16}{5}\)
\(P=5\left(x+\frac{8}{5}\right)^2+\frac{16}{5}\)
Do: \(\left(x+\frac{8}{5}\right)^2\ge0\Rightarrow5\left(x+\frac{8}{5}\right)^2+\frac{16}{5}\ge\frac{16}{5}\)
\(P_{Min}=\frac{16}{5}\Leftrightarrow x=-\frac{8}{5}\) Mà: \(y=2x+4\Rightarrow y=\frac{4}{5}\)
Thay \(x,y\) vào phương trình đề cho ta được:
\(m\left(-\frac{8}{5}\right)+\frac{4}{5}=7\)
\(\Leftrightarrow m=-\frac{31}{8}\)
Vậy nếu \(m=-\frac{31}{8}\) thì \(P\) đạt \(Min=\frac{16}{5}\)
