Giải thích các bước giải:
\(\begin{array}{l}
\left\{ \begin{array}{l}
x + my = 1\\
mx - 3my = 2m + 3
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
x = 1 - my\\
m\left( {1 - my} \right) - 3my = 2m + 3
\end{array} \right.\\
\Rightarrow - \left( {{m^2} + 3m} \right)y = 2m + 3 - m\\
\Leftrightarrow \left( {{m^2} + 3m} \right)y = - m - 3\\
+ ){m^2} + 3m = 0 \Leftrightarrow m\left( {m + 3} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
m = 0\\
m = - 3
\end{array} \right.\\
m = 0 \Rightarrow 0y = - 3\left( {VN} \right)\\
m = - 3 \Rightarrow 0y = - 0\left( {ld} \right)\\
+ ){m^2} + 3m \ne 0 \Leftrightarrow \left\{ \begin{array}{l}
m \ne - 3\\
m \ne 0
\end{array} \right.\\
\Rightarrow y = \dfrac{{ - m - 3}}{{{m^2} + 3m}} = \dfrac{{ - 1}}{m}\\
\Rightarrow x = 1 + \dfrac{1}{m} = \dfrac{{m + 1}}{m}\\
{x^2} + y = \dfrac{{{m^2} + 2m + 1}}{{{m^2}}} - \dfrac{m}{{{m^2}}}\\
= \dfrac{{{m^2} - m + 1}}{{{m^2}}} \ge 0\\
(do{m^2} > 0,{m^2} - m + 1 = {\left( {m - \dfrac{1}{2}} \right)^2} + \dfrac{3}{4} > 0)
\end{array}\)
