Đáp án: x=y=0 hoặc x=y=2015
Giải thích các bước giải:
$\begin{array}{l}
\left\{ \begin{array}{l}
\sqrt x + \sqrt {2015 - y} = \sqrt {2015} \left( 1 \right)\\
\sqrt {2015 - x} + \sqrt y = \sqrt {2015}
\end{array} \right.\\
\left( {Dkxd:0 \le x;y \le 2015} \right)\\
\Rightarrow \sqrt x + \sqrt {2015 - y} - \sqrt {2015 - x} - \sqrt y = 0\\
\Rightarrow \sqrt x - \sqrt y + \left( {\sqrt {2015 - y} - \sqrt {2015 - x} } \right) = 0\\
\Rightarrow \dfrac{{x - y}}{{\sqrt x + \sqrt y }} + \dfrac{{2015 - y - 2015 + x}}{{\sqrt {2015 - y} + \sqrt {2015 - x} }} = 0\\
\Rightarrow \left( {x - y} \right).\left( {\dfrac{1}{{\sqrt x + \sqrt y }} + \dfrac{1}{{\sqrt {2015 - y} + \sqrt {2015 - x} }}} \right) = 0\\
\Rightarrow x = y\\
Do:\dfrac{1}{{\sqrt x + \sqrt y }} + \dfrac{1}{{\sqrt {2015 - y} + \sqrt {2015 - x} }} > 0\\
Thay\,x = y\,vao\,\left( 1 \right)\\
\Rightarrow \sqrt x + \sqrt {2015 - x} = \sqrt {2015} \\
\Rightarrow x + 2\sqrt x .\sqrt {2015 - x} + 2015 - x = 2015\\
\Rightarrow 2\sqrt x .\sqrt {2015 - x} = 0\\
\Rightarrow \left[ \begin{array}{l}
x = y = 0\left( {tmdk} \right)\\
x = y = 2015\left( {tmdk} \right)
\end{array} \right.
\end{array}$
