Giải thích các bước giải:
Để hệ có nghiệm duy nhất
$\to \dfrac m2\ne\dfrac{1}{-1}\to m\ne -2$
Khi đó ta có : $mx+y+2x-y=18\to x(m+2)=18\to x=\dfrac{18}{m+2}$
$\to y=2x-11=2.\dfrac{18}{m+2}-11=\dfrac{-11m+24}{m+2}$
$\to P=x^2+y^2=(\dfrac{18}{m+2})^2+(\dfrac{-11m+24}{m+2})^2$
$\to P=\dfrac{121m^2-528m+900}{\left(2+m\right)^2}$
$\to P-\dfrac{9801}{610}=\dfrac{121m^2-528m+900}{\left(2+m\right)^2}-\dfrac{9801}{610}$
$\to P-\dfrac{9801}{610}=\dfrac{64009m^2-361284m+509796}{610\left(m+2\right)^2}$
$\to P-\dfrac{9801}{610}=\dfrac{\left(253m-714\right)^2}{610\left(m+2\right)^2}$
Mà $\dfrac{\left(253m-714\right)^2}{610\left(m+2\right)^2}\ge 0\quad\forall m\ne -2$
$\to P-\dfrac{9801}{610}\ge 0\to P\ge\dfrac{9801}{610} $
$\to Min P=\dfrac{9801}{610} $
$\to \left(253m-714\right)^2=0\to m=\dfrac{714}{253}$