Đáp án:
\(MinP = \dfrac{1}{2}\)
Giải thích các bước giải:
\(\begin{array}{l}
\left\{ \begin{array}{l}
x + 2y = 3m\\
2x - y = m
\end{array} \right.\\
\to \left\{ \begin{array}{l}
x + 2y = 3m\\
4x - 2y = 2m
\end{array} \right.\\
\to \left\{ \begin{array}{l}
5x = 5m\\
y = 2x - m
\end{array} \right.\\
\to \left\{ \begin{array}{l}
x = m\\
y = 2m - m
\end{array} \right.\\
\to x = y = m\\
P = {x^2} + {\left( {y - 1} \right)^2}\\
= {m^2} + {\left( {m - 1} \right)^2}\\
= {m^2} + {m^2} - 2m + 1\\
= 2{m^2} - 2m + 1\\
= 2{m^2} - 2.m\sqrt 2 .\dfrac{1}{{\sqrt 2 }} + \dfrac{1}{2} + \dfrac{1}{2}\\
= {\left( {m\sqrt 2 - \dfrac{1}{{\sqrt 2 }}} \right)^2} + \dfrac{1}{2}\\
Do:{\left( {m\sqrt 2 - \dfrac{1}{{\sqrt 2 }}} \right)^2} \ge 0\forall m\\
\to {\left( {m\sqrt 2 - \dfrac{1}{{\sqrt 2 }}} \right)^2} + \dfrac{1}{2} \ge \dfrac{1}{2}\\
\to MinP = \dfrac{1}{2}\\
\Leftrightarrow m\sqrt 2 - \dfrac{1}{{\sqrt 2 }} = 0\\
\to m = \dfrac{1}{2}
\end{array}\)
