Đáp án: $x+y=0$
Giải thích các bước giải:
Ta có:
$\begin{array}{l}
\left( {x + \sqrt {{x^2} + \sqrt {2010} } } \right)\left( {y + \sqrt {{y^2} + \sqrt {2010} } } \right) = \sqrt {2010} \\
\Leftrightarrow \left( {x - \sqrt {{x^2} + \sqrt {2010} } } \right)\left( {x + \sqrt {{x^2} + \sqrt {2010} } } \right)\left( {y + \sqrt {{y^2} + \sqrt {2010} } } \right) = \sqrt {2010} \left( {x - \sqrt {{x^2} + \sqrt {2010} } } \right)\\
\Leftrightarrow \left( {{x^2} - \left( {{x^2} + \sqrt {2010} } \right)} \right)\left( {y + \sqrt {{y^2} + \sqrt {2010} } } \right) = \sqrt {2010} \left( {x - \sqrt {{x^2} + \sqrt {2010} } } \right)\\
\Leftrightarrow - \sqrt {2010} \left( {y + \sqrt {{y^2} + \sqrt {2010} } } \right) = \sqrt {2010} \left( {x - \sqrt {{x^2} + \sqrt {2010} } } \right)\\
\Leftrightarrow y + \sqrt {{y^2} + \sqrt {2010} } = \sqrt {{x^2} + \sqrt {2010} } - x
\end{array}$
Chứng minh tương tự ta có: $x + \sqrt {{x^2} + \sqrt {2010} } = \sqrt {{y^2} + \sqrt {2010} } - y$
Suy ra ta có hệ sau:
$\begin{array}{l}
\left\{ \begin{array}{l}
y + \sqrt {{y^2} + \sqrt {2010} } = \sqrt {{x^2} + \sqrt {2010} } - x\\
x + \sqrt {{x^2} + \sqrt {2010} } = \sqrt {{y^2} + \sqrt {2010} } - y
\end{array} \right.\\
\Rightarrow x + y + \sqrt {{x^2} + \sqrt {2010} } + \sqrt {{y^2} + \sqrt {2010} } = \sqrt {{x^2} + \sqrt {2010} } + \sqrt {{y^2} + \sqrt {2010} } - \left( {x + y} \right)\\
\Leftrightarrow 2\left( {x + y} \right) = 0 \Leftrightarrow x + y = 0
\end{array}$
Vậy $x+y=0$
