Đáp án:
\(Min = \frac{7}{8}\)
Giải thích các bước giải:
\(\begin{array}{l}
\left\{ \begin{array}{l}
\left( {m + 1} \right)x - y = m + 1\\
x + \left( {m - 1} \right)y = 2
\end{array} \right.\\
\to \left\{ \begin{array}{l}
\left( {m + 1} \right)\left( {m - 1} \right)x - \left( {m - 1} \right)y = \left( {m + 1} \right)\left( {m - 1} \right)\\
x + \left( {m - 1} \right)y = 2
\end{array} \right.\\
\to \left\{ \begin{array}{l}
\left( {{m^2} - 1} \right)x - \left( {m - 1} \right)y = {m^2} - 1\\
x + \left( {m - 1} \right)y = 2
\end{array} \right.\\
\to \left\{ \begin{array}{l}
\left( {{m^2} - 1 + 1} \right)x = {m^2} + 1\\
y = \dfrac{{2 - x}}{{m - 1}}
\end{array} \right.\\
\to \left\{ \begin{array}{l}
x = \dfrac{{{m^2} + 1}}{{{m^2}}}\\
y = \dfrac{{2 - \dfrac{{{m^2} + 1}}{{{m^2}}}}}{{m - 1}} = \dfrac{{2{m^2} - {m^2} - 1}}{{{m^2}\left( {m - 1} \right)}}
\end{array} \right.\\
\to \left\{ \begin{array}{l}
x = \dfrac{{{m^2} + 1}}{{{m^2}}}\\
y = \dfrac{{{m^2} - 1}}{{{m^2}\left( {m - 1} \right)}} = \dfrac{{m + 1}}{{{m^2}}}
\end{array} \right.\\
S = x + y = \dfrac{{{m^2} + 1}}{{{m^2}}} + \dfrac{{m + 1}}{{{m^2}}}\\
= \dfrac{{{m^2} + 1 + m + 1}}{{{m^2}}}\\
= \dfrac{{{m^2} + m + 2}}{{{m^2}}} = 1 + \dfrac{1}{m} + \dfrac{2}{{{m^2}}}\left( 1 \right)\\
Đặt:\dfrac{1}{m} = t\\
\left( 1 \right) \to 1 + t + 2{t^2} = 2{t^2} + 2.t\sqrt 2 .\dfrac{1}{{2\sqrt 2 }} + \dfrac{1}{8} + \dfrac{7}{8}\\
= {\left( {t\sqrt 2 + \dfrac{1}{{2\sqrt 2 }}} \right)^2} + \dfrac{7}{8}\\
Do:{\left( {t\sqrt 2 + \dfrac{1}{{2\sqrt 2 }}} \right)^2} \ge 0\\
\to {\left( {t\sqrt 2 + \dfrac{1}{{2\sqrt 2 }}} \right)^2} + \dfrac{7}{8} \ge \dfrac{7}{8}\\
\to Min = \dfrac{7}{8}\\
\Leftrightarrow t\sqrt 2 + \dfrac{1}{{2\sqrt 2 }} = 0\\
\to t = - \dfrac{1}{4}\\
\to \dfrac{1}{m} = - \dfrac{1}{4}\\
\to m = - 4
\end{array}\)
