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 $x < 1$.