Đáp án đúng: C
Giải chi tiết:Ta có:
\({\left( {\sqrt {2x - 4} + \sqrt {y + 1} } \right)^2} = {\left( {\sqrt 2 .\sqrt {x - 2} + \sqrt {y + 1} } \right)^2} \le \left( {2 + 1} \right)\left( {x - 2 + y + 1} \right) = 3\left( {x + y - 1} \right),\,\,\forall x \ge 2,y \ge - 1\)\( \Rightarrow \sqrt {2x - 4} + \sqrt {y + 1} \le \sqrt {3\left( {x + y - 1} \right)} ,\,\,\forall x \ge 2,y \ge - 1\)
Theo đề bài: \(x + y - 1 = \sqrt {2x - 4} + \sqrt {y + 1} \Rightarrow x + y - 1 \le \sqrt {3\left( {x + y - 1} \right)} \Leftrightarrow 0 \le x + y - 1 \le 3 \Leftrightarrow 1 \le x + y \le 4\)
Đặt \(\sqrt {x + y} = t,\,\,t \in \left[ {1;2} \right]\), xét hàm số \(f\left( t \right) = 2016{t^4} - 2017\sqrt {5 - {t^2}} + \dfrac{{2018}}{t},\,t \in \left[ {1;2} \right]\)
\(\begin{array}{l}f'\left( t \right) = 8064{t^3} + \dfrac{{2017t}}{{\sqrt {5 - {t^2}} }} - \dfrac{{2018}}{{{t^2}}},\,t \in \left[ {1;2} \right]\\\,\,\,\,\,\,\,\,\,\,\,\, = \dfrac{{8064{t^5} - 2018}}{{{t^2}}} + \dfrac{{2017t}}{{\sqrt {5 - {t^2}} }} > 0,\forall t \in \left[ {1;2} \right]\end{array}\)
\( \Rightarrow \) Hàm số \(f\left( t \right)\) đồng biến trên \(\left[ {1;2} \right]\) \( \Rightarrow \left\{ \begin{array}{l}\mathop {\min }\limits_{t \in \left[ {1;2} \right]} f\left( t \right) = f\left( 1 \right) = 0\\\mathop {\max }\limits_{t \in \left[ {1;2} \right]} f\left( t \right) = f\left( 2 \right) = 31248\end{array} \right.\)
Vậy \({S_{\min }} = 0\) khi và chỉ khi \(\left\{ \begin{array}{l}x + y = 1\\x + y - 1 = \sqrt {2x - 4} + \sqrt {y + 1} \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}x + y = 1\\\sqrt {2x - 4} + \sqrt {y + 1} = 0\end{array} \right.\)\( \Leftrightarrow \left\{ \begin{array}{l}x = 2\\y = - 1\end{array} \right.\)
\({S_{\max }} = 31248\) khi và chỉ khi \(\left\{ \begin{array}{l}x + y = 2\\x + y - 1 = \sqrt {2x - 4} + \sqrt {y + 1} \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}x + y = 2\\\sqrt {2x - 4} + \sqrt {y + 1} = 1\end{array} \right.\)\(\begin{array}{l} \Leftrightarrow \left\{ \begin{array}{l}x = 2 - y\\\sqrt { - 2y} + \sqrt {y + 1} = 1\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}x = 2 - y\\ - 2y + 2\sqrt { - 2y\left( {y + 1} \right)} + y + 1 = 1\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}x = 2 - y\\2\sqrt { - 2y\left( {y + 1} \right)} = y\end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l}x = 2 - y\\ - 8{y^2} - 8y = {y^2}\\y \ge 0\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}x = 2 - y\\\left[ \begin{array}{l}y = 0\\y = - \dfrac{8}{9}\end{array} \right.\\y \ge 0\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}x = 2 - y\\y = 0\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}x = 2\\y = 0\end{array} \right..\end{array}\)
