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Question Number 45346 by Tawa1 last updated on 12/Oct/18

Commented by Meritguide1234 last updated on 12/Oct/18

$$thisismyoldpost45204$$

Commented by Tawa1 last updated on 12/Oct/18

$$\mathrm{What}\mathrm{is}\mathrm{the}\mathrm{solution}$$

Answered by tanmay.chaudhury50@gmail.com last updated on 12/Oct/18

$$=\underset{r=1}{\sum ^{99}}\frac{\sqrt{10+\sqrt{r}}}{\sqrt{10-\sqrt{r}}}$$$$\sqrt{10+\sqrt{99}}>\sqrt{10+\sqrt{1}}>\sqrt{10+\sqrt{1}}$$$$\sqrt{10+\sqrt{99}}>\sqrt{10+\sqrt{2}}>\sqrt{10+\sqrt{1}}$$$$\sqrt{10+\sqrt{99}}>\sqrt{10+\sqrt{3}}>\sqrt{10+\sqrt{1}}$$$$...........$$$$...........addingthem$$$$$$$$99\times \sqrt{10+\sqrt{99}}>N_r>99\times \sqrt{11}$$$$x=10+\sqrt{99}$$$$2x=20+2\times \sqrt{11}\times 3$$$$2x={(\sqrt{11}+3)}^2$$$$x=\frac{1}{2}{(\sqrt{11}+3)}^2$$$$\sqrt{x}=\frac{3+\sqrt{11}}{\sqrt{2}}=(\frac{3}{\sqrt{2}}+\sqrt{\frac{11}{2}})$$$$99\times (\frac{3}{\sqrt{2}}+\sqrt{\frac{11}{2}})>N_r>99\sqrt{11}$$$$itislogicbasedjustification...microsoft$$$$Excellcangivetheansweroftheproblem$$$$\sqrt{10-\sqrt{99}}<\sqrt{10-\sqrt{1}}<\sqrt{10-\sqrt{1}}$$$$\sqrt{10-\sqrt{99}}<\sqrt{10-\sqrt{2}}<\sqrt{10-\sqrt{1}}$$$$\sqrt{10-\sqrt{99}}<\sqrt{10-\sqrt{3}}<\sqrt{10-\sqrt{1}}$$$$........$$$$........addthem$$$$99\times \sqrt{10-\sqrt{99}}<D_r<99\times 3$$$$99\times (\sqrt{\frac{11}{2}}-\frac{3}{\sqrt{2}})<D_r<99\times 3$$

Commented by Tawa1 last updated on 12/Oct/18

$$\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}.\mathrm{What}\mathrm{is}\mathrm{the}\mathrm{final}\mathrm{answer}.$$

Commented by Meritguide1234 last updated on 12/Oct/18

$$ansis\sqrt{2}+1$$

Commented by Meritguide1234 last updated on 12/Oct/18

Commented by Tawa1 last updated on 12/Oct/18

$$\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}$$

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