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?

The domain gives $-1\leq x<0$ or $x\geq1$.


*

*$-1\leq x<0$.


We obtain:
$$\sqrt{\frac{(1-x)(x+1)}{-x}}=\frac{\left(\sqrt{1-x}\right)^2}{x}+\sqrt{\frac{1-x}{-x}}$$ or
$$\sqrt{\frac{x+1}{-x}}=\frac{\sqrt{1-x}}{x}+\frac{1}{\sqrt{-x}}$$ or
$$-\sqrt{-x(x+1)}=\sqrt{1-x}-\sqrt{-x}$$ or
$$\sqrt{-x}=\sqrt{1-x}+\sqrt{-x(1+x)}$$ or
$$-x=1-x-x-x^2+2\sqrt{x(x^2-1)}$$ or
$$x^2+x-1=2\sqrt{x(x^2-1)},$$ which is impossible because $x^2+x-1<0$ for $-1\leq x<0.$


*$x\geq1$.


We obtain:
$$\sqrt{\frac{(x-1)(x+1)}{x}}=\frac{x-1}{x}+\sqrt{\frac{x-1}{x}},$$ which gives
$$x=1$$ or
$$\sqrt{\frac{x+1}{x}}=\frac{\sqrt{x-1}}{x}+\sqrt{\frac{1}{x}},$$ which is
$$\sqrt{(x+1)x}=\sqrt{x-1}+\sqrt{x}$$ or
$$x^2+x=2x-1+2\sqrt{(x-1)x}$$ or
$$x^2-x-2\sqrt{(x-1)x}+1=0$$ or
$$\left(\sqrt{x^2-x}-1\right)^2=0$$ or
$$x^2-x-1=0,$$ which gives
$$x=\frac{1+\sqrt{5}}{2}$$ and we got the answer:
$$\left\{1,\frac{1+\sqrt5}{2}\right\}$$
