Đáp án:
\(Min = - 1\)
Giải thích các bước giải:
\(\begin{array}{l}
\left\{ \begin{array}{l}
x + my = m + 1\\
mx + y = 3m - 1
\end{array} \right.\\
\to \left\{ \begin{array}{l}
- mx - {m^2}y = - {m^2} - m\\
mx + y = 3m - 1
\end{array} \right.\\
\to \left\{ \begin{array}{l}
\left( {1 - {m^2}} \right)y = - {m^2} + 2m - 1\\
x = m + 1 - my
\end{array} \right.\\
\to \left\{ \begin{array}{l}
y = \dfrac{{ - {{\left( {m - 1} \right)}^2}}}{{\left( {1 - m} \right)\left( {m + 1} \right)}}\\
x = m + 1 - my
\end{array} \right.\\
\to \left\{ \begin{array}{l}
y = \dfrac{{m - 1}}{{m + 1}}\\
x = m + 1 - m.\dfrac{{m - 1}}{{m + 1}}
\end{array} \right.\\
\to \left\{ \begin{array}{l}
y = \dfrac{{m - 1}}{{m + 1}}\\
x = \dfrac{{{m^2} + 2m + 1 - {m^2} + m}}{{m + 1}}
\end{array} \right.\\
\to \left\{ \begin{array}{l}
y = \dfrac{{m - 1}}{{m + 1}}\\
x = \dfrac{{3m + 1}}{{m + 1}}
\end{array} \right.\\
DK:m \ne - 1\\
Có:x.y = \dfrac{{3m + 1}}{{m + 1}}.\dfrac{{m - 1}}{{m + 1}}\\
= \dfrac{{\left( {3m + 1} \right)\left( {m - 1} \right)}}{{{{\left( {m + 1} \right)}^2}}}\\
= \dfrac{{3{m^2} - 3m + m - 1}}{{{{\left( {m + 1} \right)}^2}}}\\
= \dfrac{{3{m^2} - 2m - 1}}{{{{\left( {m + 1} \right)}^2}}} = \dfrac{{3\left( {{m^2} + 2m + 1} \right) - 8m - 4}}{{{{\left( {m + 1} \right)}^2}}}\\
= \dfrac{{3{{\left( {m + 1} \right)}^2} - 8\left( {m + 1} \right) + 4}}{{{{\left( {m + 1} \right)}^2}}}\\
= 3 - \dfrac{8}{{m + 1}} + \dfrac{4}{{{{\left( {m + 1} \right)}^2}}}\left( 1 \right)\\
Đặt:\dfrac{1}{{\left( {m + 1} \right)}} = t\\
\left( 1 \right) \to 4{t^2} - 8t + 3 = 4\left( {{t^2} - 2t} \right) + 3\\
= 4\left( {{t^2} - 2t + 1 - 1} \right) + 3\\
= 4{\left( {t - 1} \right)^2} - 1\\
Do:{\left( {t - 1} \right)^2} \ge 0\forall t\\
\to 4{\left( {t - 1} \right)^2} \ge 0\\
\to 4{\left( {t - 1} \right)^2} - 1 \ge - 1\\
\to Min = - 1\\
\Leftrightarrow t - 1 = 0\\
\to t = 1\\
\to \dfrac{1}{{\left( {m + 1} \right)}} = 1\\
\to m + 1 = 1\\
\to m = 0
\end{array}\)
