Solve: $\sqrt{x-\frac1x}-\sqrt{1-\frac1x}=1-\frac1x$ for $ x\neq0$ $$\sqrt{x-\frac{1}{x}}-\sqrt{1-\frac{1}{x}}{=1-\frac{1}{x}}, x\neq0$$
So my idea is that I moved the second sqrt to the RHS and left the first on the LHS. I then squared everything and ended up with $\frac{x^2-1}{x}=2-\frac{3}{x}+\frac{1}{x^2}$. I don't know what to do next? Can somebody give a hint?

Note that $x=1$ is a solution. Else, let $A = \sqrt{x-\frac{1}{x}}$ and $B = \sqrt{1-\frac{1}{x}}$
then $A-B = \frac{x-1}{x}$ and $A^2-B^2 = x-1$, then dividing we get $A+B = x$
Adding we get $2\sqrt{x-\frac{1}{x}} = 1+x - \frac{1}{x}$ or $2A = 1+A^2 \Rightarrow A=1$ or $x^2-x+1 = 0$
