Đáp án:
\(Max = \dfrac{7}{8}\)
Giải thích các bước giải:
\(\begin{array}{l}
\left\{ \begin{array}{l}
\left( {m + 1} \right)\left( {m - 1} \right)x - \left( {m - 1} \right)y = \left( {m + 1} \right)\left( {m - 1} \right)\\
x + \left( {m - 1} \right)y = 2
\end{array} \right.\\
\to \left\{ \begin{array}{l}
\left( {{m^2} - 1 + 1} \right)x = {m^2} - 1 + 2\\
y = \left( {m + 1} \right)x - m - 1
\end{array} \right.\\
\to \left\{ \begin{array}{l}
x = \dfrac{{{m^2} + 1}}{{{m^2}}}\\
y = \left( {m + 1} \right).\dfrac{{{m^2} + 1}}{{{m^2}}} - m - 1
\end{array} \right.\\
\to \left\{ \begin{array}{l}
x = \dfrac{{{m^2} + 1}}{{{m^2}}}\\
y = \dfrac{{{m^3} + m + {m^2} + 1 - {m^3} - {m^2}}}{{{m^2}}}
\end{array} \right.\\
\to \left\{ \begin{array}{l}
x = \dfrac{{{m^2} + 1}}{{{m^2}}}\\
y = \dfrac{{m + 1}}{{{m^2}}}
\end{array} \right.\\
DK:m \ne 0\\
Có:S = x + y = \dfrac{{{m^2} + 1}}{{{m^2}}} + \dfrac{{m + 1}}{{{m^2}}}\\
= \dfrac{{{m^2} + m + 2}}{{{m^2}}}\\
= 1 + \dfrac{1}{m} + \dfrac{2}{{{m^2}}}\\
Đặt:\dfrac{1}{m} = t \to \dfrac{1}{{{m^2}}} = {t^2}\\
S = 1 + t + 2{t^2}\\
= 2{t^2} + 2.t\sqrt 2 .\dfrac{1}{{2\sqrt 2 }} + \dfrac{1}{8} + \dfrac{7}{8}\\
= {\left( {t\sqrt 2 + \dfrac{1}{{2\sqrt 2 }}} \right)^2} + \dfrac{7}{8}\\
Do:{\left( {t\sqrt 2 + \dfrac{1}{{2\sqrt 2 }}} \right)^2} \ge 0\forall t\\
\to {\left( {t\sqrt 2 + \dfrac{1}{{2\sqrt 2 }}} \right)^2} + \dfrac{7}{8} \ge \dfrac{7}{8}\\
\to Max = \dfrac{7}{8}\\
\Leftrightarrow t\sqrt 2 + \dfrac{1}{{2\sqrt 2 }} = 0\\
\to t = - \dfrac{1}{4}\\
\to x = - 4
\end{array}\)
