55rd Ukrainian National Mathematical Olympiad - Fourth Round

Multiply both numerator and denominator by the conjugate: $$\frac{\sqrt{2014+\sqrt{x}}}{2014-x}+\frac{\sqrt{2015+\sqrt{x+1}}}{2015-(x+1)}+\frac{\sqrt{2016+\sqrt{x+2}}}{2016-(x+2)}=\frac{\sqrt{2015-x+\sqrt{1}}}{(2015-x)-1}+\frac{\sqrt{2016-x+\sqrt{2}}}{(2016-x)-2}+\frac{\sqrt{2017-x+\sqrt{3}}}{(2017-x)-3}$$ After multiplying by the common denominator we have: $$\sqrt{2014+\sqrt{x}}+\sqrt{2015+\sqrt{x+1}}+\sqrt{2016+\sqrt{x+2}}=\sqrt{2015-x}+\sqrt{1}+\sqrt{2016-x}+\sqrt{2}+\sqrt{2017-x}+\sqrt{3},$$

1) $x=1$ is the solution of the equation: $$\sqrt{2014}+\sqrt{1}+\sqrt{2015}+\sqrt{2}+\sqrt{2016}+\sqrt{3}=\sqrt{2014}+\sqrt{1}+\sqrt{2015}+\sqrt{2}+\sqrt{2016}+\sqrt{3}.$$

2) Denote $S=\sqrt{2014}+\sqrt{1}+\sqrt{2015}+\sqrt{2}+\sqrt{2016}+\sqrt{3}$. For $x>1$ we have $$\sqrt{2014}+\sqrt{x}+\sqrt{2015}+\sqrt{x+1}+\sqrt{2016}+\sqrt{x+2}>\sqrt{2014}+\sqrt{1}+\sqrt{2015}+\sqrt{2}+\sqrt{2016}+\sqrt{3}=S.$$ So the left-hand side is greater than $S$, and in the same way we analyze the right-hand side: $$\sqrt{2015-x}+\sqrt{1}+\sqrt{2016-x}+\sqrt{2}+\sqrt{2017-x}+\sqrt{3}<\sqrt{2014}+\sqrt{1}+\sqrt{2015}+\sqrt{2}+\sqrt{2016}+\sqrt{3}=S.$$ So in this interval there are no roots. The same holds for $x<1$.