Có: \(\left\{{}\begin{matrix}x-2y=3-m\\2x+y=3\left(m+2\right)\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=3-m+2y\left(1\right)\\2x=3m+6-y\left(2\right)\end{matrix}\right.\)
Thay (1) vào (2) ta có:
\(2\left(3-m+2y\right)=3m+6-y\)
\(\Leftrightarrow5m-5y=0\)
\(\Leftrightarrow y=m\)
\(\Rightarrow x=3-m+2m=3+m\)
Thay x = 3 + m, y = m vào A, được:
\(A=\left(3+m\right)^2+m^2+4\)
\(A=2m^2+6m+13\)
\(A=2\left(m^2+2.\dfrac{3}{2}m+\dfrac{9}{4}\right)+\dfrac{17}{2}\)
\(A=2\left(m+\dfrac{3}{2}\right)^2+\dfrac{17}{2}\ge\dfrac{17}{2}\)
Dấu "=" xảy ra <=> \(m+\dfrac{3}{2}=0\Leftrightarrow m=-\dfrac{3}{2}\)
Vậy minA = \(\dfrac{17}{2}\) <=> m = \(-\dfrac{3}{2}\).
