Equation with radicals and reciprocals Find all $x\in\mathbb{R}$ satisfying $$x=\sqrt{x-\frac{1}{x}}+\sqrt{1-\frac{1}{x}}.$$

Multiply both sides by $x^{1/2}$ to get $$x^{3/2} = \sqrt{x^2-1} + \sqrt{x-1}.$$Making the substitution $a = \sqrt{x^2 - 1}$, $b=\sqrt{x-1}$, we have $a+b = x^{3/2}$ and $a^2 - b^2 = x^2 - x$, so $a-b = (a^2-b^2)/(a+b) = \frac{x-1}{x^{1/2}}$. Then solving for $b$ gives $b = \frac{x^{3/2} - x^{1/2} + x^{-1/2}}{2}$. Then, we get the equation $$\frac{x^{3/2} - x^{1/2} + x^{-1/2}}{2} = \sqrt{x-1},$$which we can expand/rearrange to get $$x^4 - 2x^3 - x^2 + 2x + 1 = 0,$$which factors as $$\left(x^2 - x - 1\right)^2 = 0 \implies x^2 - x - 1 = 0,$$which has solutions $x = \frac{1\pm\sqrt{5}}{2}$, but note that we must have $x \ge 1$ by the problem, so $\boxed{x = \frac{1+\sqrt{5}}{2}}$.
I would be interested in seeing a slicker solution, though. 

Here is a slicker solution I found.
Use David's substitution $y=\sqrt{x-\frac{1}{x}}, z=\sqrt{1-\frac{1}{x}}$, and note that $x=y^2-z^2+1$. Then multiply $x=y+z\;(\star)$ by $y-z$ to get $$y-z=\frac{y^2-z^2}{x}=z^2.$$ Summing this with $(\star)$ we get $$x-2y+z^2=0\iff(y-1)^2=0.$$ Thus $y=1$, and $x\geq1$ is the root of $x^2-x-1$, so $\boxed{x = \frac{1+\sqrt{5}}{2}}$.
