Solution to Equation

$\begin{aligned} x & = \sqrt{x-\frac 1x} + \sqrt{1-\frac 1x} & \small \color{#3D99F6} \text{Multiply both sides by }\sqrt x \\ x\sqrt x & = \sqrt{x^2-1} + \sqrt{x-1} \\ & = \sqrt{x-1}\left(\sqrt{x+1}+1\right) & \small \color{#3D99F6} \text{Squaring both sides} \\ x^3 & = (x-1)\left(x+1+2\sqrt{x+1}+1\right) \\ x^3 & = x^2+x-2+2\sqrt{(x+1)(x^2-2x+1)} \\ x^3-x^2-x+2 & = 2\sqrt{x^3-x^2-x+1} & \small \color{#3D99F6} \text{Let }u = \sqrt{x^3-x^2-x+1} \\ \implies u^2 - 2u + 1 & = 0 \\ (u-1)^2 & = 0 \\ u & = 1 \\ \sqrt{x^3-x^2-x+1} & = 1 \\ x^3-x^2-x+1 & = 1 \\ x(x^2 - x-1) & = 0 \end{aligned}$

$\implies \begin{cases} x = 0 \\ x = \dfrac {1+\sqrt 5}2 \\ x = \dfrac {1-\sqrt 5}2 \end{cases}$

But only $x = \dfrac {1+\sqrt 5}2 = \varphi$ , the golden ratio, satisfies the original equation. Therefore the sum of all real value of $x$ satisfying the equation is $\varphi \approx \boxed{1.618}$ .

Since (x-1/x)^1/2 and (1-1/x)^1/2 both are greater equal to 0 then x is greater equal to 0. As 1/x to be defined , precisely x greater than 0. Now squaring both side of the equation and multiplying both side by x followed by rearrangement gives x^3 - x^2 - x + 2 = 2×(x^3 - x^2 - x + 1)^1/2 Which leads to x^2 - x - 1 = 0 So x may be (1+√5)/2 or (1-√5)/2 But positive value is the only acceptable value here. Again checking the uniqueness of the solution it gets to be unique. So x= (1+√5)/2