ĐK: \(3\le x\le5\)
\(\begin{align} & VT=\left( \sqrt{x-3}+\sqrt{5-x} \right)\le 2\left( x-3+5-x \right) \\ & \Leftrightarrow {{\left( \sqrt{x-3}+\sqrt{5-x} \right)}^{2}}\le 4 \\ & \Rightarrow \sqrt{x-3}+\sqrt{5-x}\le 2 \\ & VP={{\left( y+\sqrt{2013} \right)}^{2}}+2\ge 2 \\ \end{align}\)
Vậy phương trình chỉ tồn tại khi $VT=VP=2$
\(\Leftrightarrow\left\{{}\begin{matrix}\left(\sqrt{x-3}+\sqrt{5-x}\right)^2=2^2\\\left(y+\sqrt{2013}\right)^2+2=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=4\\y=\sqrt{2013}\end{matrix}\right.\)
