- Đk: \(\left\{{}\begin{matrix}x-2012\ge0\\y-2013\ge0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge2012\\y\ge2013\end{matrix}\right.\)
\(\dfrac{16}{\sqrt{x-2012}}+\dfrac{1}{\sqrt{y-2013}}=10-\sqrt{x-2012}-\sqrt{y-2013}\)
\(\Leftrightarrow\dfrac{16}{\sqrt{x-2012}}-8+\sqrt{x-2012}+\dfrac{1}{\sqrt{y-2013}}-2+\sqrt{y-2013}=0\)
\(\Leftrightarrow\left(\dfrac{4}{\sqrt[4]{x-2012}}-\sqrt[4]{x-2012}\right)^2+\left(\dfrac{1}{\sqrt[4]{y-2013}}-\sqrt[4]{y-2013}\right)^2=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{4}{\sqrt[4]{x-2012}}=\sqrt[4]{x-2012}\\\dfrac{1}{\sqrt[4]{y-2013}}=\sqrt[4]{y-2013}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{x-2012}=4\\\sqrt{y-2013}=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x-2012=16\\y-2013=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=2028\\y=2014\end{matrix}\right.\) (nhận)
- Vậy phương trình có nghiệm \(\left(2028;2014\right)\)
