<s>[INST] Find the positive value of $x$ satisfying the given equation $${\sqrt {x^2- 1\over x}} + {\sqrt{x-1\over x}} = x$$
Tried removing roots. Got a degree $6$ equation which I didn't no how to solve. Also tried substituting $x = \sec(y)$ but couldn't even come close to the solution.
 [/INST] Checking other answers:
Note that $\frac{x^2-1}{x}\ge 0;\frac{x-1}{x}\ge 0$ and $x>0$ imply $x>1$. 
If $x^2=x+1 \ \ (1)$, then:
$$ x^2-1=x \Rightarrow \frac{x-1}{x}=\frac{1}{x+1},\\
{\sqrt {x^2- 1\over x}} + {\sqrt{x-1\over x}} = x \ \ \ \ \ \ \ \ \ \ (2) \ \ \ \ \ \Rightarrow \\
{\sqrt {(x+1)- 1\over x}} + {\sqrt{1\over x+1}} = \sqrt{x+1} \Rightarrow \\
1+\frac1{\sqrt{x+1}}=\sqrt{x+1} \Rightarrow \\
\sqrt{x+1}+1=x+1 \Rightarrow \\
x+1=x^2.$$
So, $(1)$ and $(2)$ are equivalent.
</s>