\(\dfrac{\sqrt{x-2015}-1}{x-2015}+\dfrac{\sqrt{y-2016}-1}{y-2016}=\dfrac{1}{2}\)
Điều kiện \(\left\{{}\begin{matrix}x>2015\\y>2016\end{matrix}\right.\)
\(\Leftrightarrow\dfrac{1}{\sqrt{x-2015}}-\dfrac{1}{x-2015}+\dfrac{1}{\sqrt{y-2016}}-\dfrac{1}{y-2016}=\dfrac{1}{2}\)
Đặt \(\left\{{}\begin{matrix}\dfrac{1}{\sqrt{x-2015}}=a>0\\\dfrac{1}{\sqrt{y-2016}}=b>0\end{matrix}\right.\) thì ta có:
\(a-a^2+b-b^2=\dfrac{1}{2}\)
\(\Leftrightarrow\left(2a^2-2a+\dfrac{1}{2}\right)+\left(2b^2-2b+\dfrac{1}{2}\right)=0\)
\(\Leftrightarrow\left(\sqrt{2}a-\dfrac{1}{\sqrt{2}}\right)^2+\left(\sqrt{2}b-\dfrac{1}{\sqrt{2}}\right)^2=0\)
\(\Leftrightarrow a=b=\dfrac{1}{4}\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{1}{\sqrt{x-2015}}=\dfrac{1}{4}\\\dfrac{1}{\sqrt{y-2016}}=\dfrac{1}{4}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=2019\\y=2020\end{matrix}\right.\)