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Question Number 132292 by Algoritm last updated on 13/Feb/21

Answered by mathmax by abdo last updated on 13/Feb/21

$$\frac{1}{\left[\mathrm{x}\right]}+\frac{1}{\left[2\mathrm{x}\right]}=\left[\mathrm{x}\right]+\frac{1}{3}\mathrm{let}\left[\mathrm{x}\right]=\mathrm{n}\Rightarrow \mathrm{n}⩽\mathrm{x}<\mathrm{n}+1\Rightarrow 2\mathrm{n}⩽2\mathrm{x}<2\mathrm{n}+2$$$$\mathrm{if}2\mathrm{x}\in [2\mathrm{n},2\mathrm{n}+1[\Rightarrow \mathrm{x}\in [\mathrm{n},\mathrm{n}+\frac{1}{2}[\Rightarrow \left[2\mathrm{x}\right]=2\mathrm{n}$$$$\mathrm{e}\Rightarrow \frac{1}{\mathrm{n}}+\frac{1}{2\mathrm{n}}=\mathrm{n}+\frac{1}{3}\Rightarrow \frac{3}{2\mathrm{n}}=\mathrm{n}+\frac{1}{3}\Rightarrow \frac{9}{2\mathrm{n}}=3\mathrm{n}+1\Rightarrow 9=2\mathrm{n}(3\mathrm{n}+1)\Rightarrow$$$${6\mathrm{n}}^2+2\mathrm{n}-9=0\to {\mathrm{\Delta }}^{{}^{\prime }}=1-6.(-9)=1+54=55\Rightarrow {\mathrm{n}}_1=\frac{-1+\sqrt{54}}{12}$$$$\mathrm{not}\mathrm{solution}\mathrm{also}{\mathrm{n}}_2=\frac{-1-\sqrt{54}}{12}\mathrm{not}\mathrm{solution}(\mathrm{n}\mathrm{integr}!)$$$$\mathrm{if}2\mathrm{x}\in [2\mathrm{n}+1,2\mathrm{n}+2[\Rightarrow \mathrm{x}\in [\mathrm{n}+\frac{1}{2},\mathrm{n}+1[\left[2\mathrm{x}\right]=2\mathrm{n}+1$$$$\mathrm{e}\Rightarrow \frac{1}{\mathrm{n}}+\frac{1}{2\mathrm{n}+1}=\mathrm{n}+\frac{1}{3}\Rightarrow \frac{3\mathrm{n}+1}{\mathrm{n}(2\mathrm{n}+1)}=\frac{3\mathrm{n}+1}{3}\Rightarrow$$$${2\mathrm{n}}^2+\mathrm{n}=3\Rightarrow {2\mathrm{n}}^2+\mathrm{n}-3=0$$$$\mathrm{\Delta }=1-4\left(2\right)(-3)=1+24=25\Rightarrow {\mathrm{n}}_1=\frac{-1+5}{4}=1\mathrm{and}{\mathrm{n}}_2=\frac{-1-5}{4}=-\frac{3}{2}$$$$\mathrm{not}\mathrm{solution}\Rightarrow \mathrm{n}=1\Rightarrow \mathrm{x}\in [\frac{3}{2},2[\mathrm{this}\mathrm{is}\mathrm{the}\mathrm{set}\mathrm{of}\mathrm{solution}$$

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