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Question Number 166391 by mnjuly1970 last updated on 19/Feb/22

$$$$$$solvein\mathbb{R}$$$$\sqrt{x.\lfloor x\rfloor }-\sqrt{\lfloor x\rfloor }=1$$$$-------$$

Answered by mr W last updated on 19/Feb/22

$$sayx=n+f$$$$1=\sqrt{n(n+f)}-\sqrt{n}⩾n-\sqrt{n}$$$${\left(\sqrt{n}\right)}^2-\sqrt{n}-1⩽0$$$$\sqrt{n}⩽\frac{1+\sqrt{5}}{2}\Rightarrow n⩽\frac{3+\sqrt{5}}{2}$$$$\Rightarrow n⩽2...\left(i\right)$$$$1=\sqrt{n(n+f)}-\sqrt{n}<\sqrt{n(n+1)}-\sqrt{n}$$$$\Rightarrow n⩾2...\left(ii\right)$$$$\Rightarrow n=2$$$$\sqrt{2x}-\sqrt{2}=1$$$$2x={(1+\sqrt{2})}^2=3+2\sqrt{2}$$$$\Rightarrow x=\frac{3}{2}+\sqrt{2}$$

Commented by mnjuly1970 last updated on 19/Feb/22

$$bravosirW$$

Answered by mnjuly1970 last updated on 19/Feb/22

$$\sqrt{\lfloor x\rfloor }(\sqrt{x}-1)=1\mathrm{D}=(1,\mathrm{\infty })$$$$\{\sqrt{x}-1=\frac{1}{\sqrt{\lfloor x\rfloor }}\}\lfloor x\rfloor =n$$$$\sqrt{x}⩾\sqrt{\lfloor x\rfloor }$$$$\sqrt{n}-1⩽\frac{1}{\sqrt{n}}\Rightarrow n-\sqrt{n}⩽1$$$$n^2-2n+1⩽n$$$$n^2-3n+1⩽0$$$${(n-\frac{3}{2})}^2⩽\frac{5}{4}$$$$n⩽\frac{\sqrt{5}+3}{2}\Rightarrow n=2$$$$\sqrt{x}=1+\frac{1}{\sqrt{2}}\Rightarrow x=\frac{3+2\sqrt{2}}{2}$$$$$$

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