<s>[INST] A golden question $\sqrt{2+\sqrt{2-x}}=\sqrt{x-\frac 1x} + \sqrt{1-\frac 1x}$ How would you solve this problem for real $x$?
$$\sqrt{2+\sqrt{2-x}}=\sqrt{x-\frac 1x} + \sqrt{1-\frac 1x}$$

It can be easily shown that both equations
$$x=\sqrt{2+\sqrt{2-x}}\tag{1}$$
and $$x=\sqrt{x-\frac 1x} + \sqrt{1-\frac 1x}\tag{2}$$
have the same real solution, i.e. the golden ratio, $$x = \frac {\sqrt 5 + 1}2=\phi.$$
The main question is derived by combining (1) and (2) into one equation. The final solution is still the same, but the approach might not be as straightforward.
 [/INST] (The following is about the other roots in ConMan's answer, but it was too long for a comment.)
Courtesy WA, the original equation can be brought to polynomial form with groebnerBasis[ {u - v - w, x - 2 + (u^2 - 2)^2, v^2 x - x^2 + 1, w^2 x - x + 1}, {x}, {u, v, w}  ] which returns and factors:
$$
x^6 - 10 x^5 + 39 x^4 - 22 x^3 - 63 x^2 + 32 x + 32 = (x^2 - x - 1) (x^4 - 9 x^3 + 31 x^2 - 32)
$$

*

*The positive root $\,\varphi\,$ of the quadratic and the negative real root $\,\simeq -0.8959\,$ of the quartic satisfy the original equation $\,\sqrt{2 + \sqrt{2-x}} = x = \sqrt{x - \frac{1}{x}} + \sqrt{1 - \frac{1}{x}}\,$.


*The negative root $\,-\dfrac{1}{\varphi}\,$ of the quadratic satisfies $\,\sqrt{2 - \sqrt{2-x}} = -x = - \sqrt{x - \frac{1}{x}} + \sqrt{1 - \frac{1}{x}}\,$.


*The positive real root $\,\simeq 1.2197\,$ of the quartic satisfies $\,\sqrt{2 - \sqrt{2-x}} = \sqrt{x - \frac{1}{x}} + \sqrt{1 - \frac{1}{x}}\,$.


*The two complex roots $\,\simeq 4.3381 \pm 3.2349 i\,$ of the quartic satisfy $\,\sqrt{2 - \sqrt{2-x}} = \sqrt{x - \frac{1}{x}} - \sqrt{1 - \frac{1}{x}}\,$.
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