solve equation

[requferofile]

solve in real numbers

$\frac{\lfloor 2x\rfloor}{\lfloor x\rfloor}=\lfloor x\rfloor+\frac{1}{\lfloor x\rfloor}$

[mihiv]

$1\leqslant x<\frac 32,2.5\leqslant x<3$

[ShMaxG]

$\[x \in \left[ {1,\frac{3} {2}} \right) \cup \left[ {\frac{5} {2},3} \right)\] $

[ShMaxG]

Here one can just consider some cases:
1) $\[x \in \left( { - 1;1} \right)\]$;
2) $\[x \leqslant - 1\]$;
3) $\[x \in \left[ {1;1 + \frac{1} {2}} \right) \cup \left[ {2;2 + \frac{1} {2}} \right) \cup ...\]$. Here $\[\left\lfloor {2x} \right\rfloor = 2\left\lfloor x \right\rfloor \]$;
4)$ \[x \in \left[ {1 + \frac{1} {2};2} \right) \cup \left[ {2 + \frac{1} {2};3} \right) \cup ...\]$. Here $\[\left\lfloor {2x} \right\rfloor = 2\left\lfloor x \right\rfloor + 1\]$.

That's all.