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Question Number 191343 by mnjuly1970 last updated on 23/Apr/23

$$$$$$\mathrm{solve}in\mathbb{R}$$$$$$$$\lfloor \frac{1}{x}\rfloor +\lfloor \frac{2}{x}\rfloor +\lfloor \frac{3}{x}\rfloor =1$$$$$$

Answered by mr W last updated on 23/Apr/23

$$\lfloor \frac{3}{x}\rfloor =1and\lfloor \frac{2}{x}\rfloor =0$$$$\Rightarrow 1⩽\frac{3}{x}<2\Rightarrow \frac{3}{2}<x⩽3and$$$$\Rightarrow \frac{2}{x}<1\Rightarrow x>2$$$$solutionis2<x⩽3✓$$

Commented by mnjuly1970 last updated on 23/Apr/23

$$thankyousomuchsirW\underset{―}{\underbrace{⋚}}$$

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