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Question Number 121138 by Ar Brandon last updated on 05/Nov/20

Answered by TANMAY PANACEA last updated on 05/Nov/20

$$x^{x-x^{-\frac{1}{2}}}={\left(x^{-\frac{1}{2}}\right)}^{x+x^{\frac{-1}{2}}}$$$$x-x^{-\frac{1}{2}}=\frac{-x}{2}-\frac{1}{2}{x}^{-\frac{1}{2}}$$$$\frac{3x}{2}=\frac{1}{2}{x}^{\frac{-1}{2}}$$$$3x=x^{\frac{-1}{2}}$$$$3x^{\frac{3}{2}}=1$$$$x^{\frac{3}{2}}=\frac{1}{3}\to x={\left(\frac{1}{3}\right)}^{\frac{2}{3}}=\frac{1}{\sqrt[3]{9}}$$$$y=\frac{1}{\sqrt{x}}=\frac{1}{{\left(\frac{1}{3}\right)}^{\frac{2}{3}\times \frac{1}{2}}}=\frac{1}{\frac{1}{\sqrt[3]{3}}}=\sqrt[)]{3}$$$$optionD(\frac{1}{\sqrt[3]{9}},\sqrt[3]{3})$$

Commented by Ar Brandon last updated on 05/Nov/20

Thanks Sir

Commented by TANMAY PANACEA last updated on 05/Nov/20

$$mostwelcomesir$$

Answered by Dwaipayan Shikari last updated on 05/Nov/20

$$x^{x-y}=y^{x+y}$$$$\Rightarrow x^{x+y}=y^{x+y}{x}^{2y}$$$$\Rightarrow {\left(\frac{x}{y}\right)}^{x+y}=x^{2y}$$$$\Rightarrow {\left(x^{\frac{3}{2}}\right)}^{x+\frac{1}{\sqrt{x}}}=x^{\frac{2}{\sqrt{x}}}$$$$\Rightarrow x=1$$$$3(x+\frac{1}{\sqrt{x}})=\frac{4}{\sqrt{x}}$$$$\Rightarrow 3x=\frac{1}{\sqrt{x}}\Rightarrow x=\frac{1}{\sqrt[3]{9}}$$$$y=\sqrt[3]{3}$$$$Sox=1,y=1x=\frac{1}{\sqrt[3]{9}}y=\sqrt[3]{3}$$

Commented by Ar Brandon last updated on 05/Nov/20

Brilliant ! Thanks Mr Dwaipayan����

Commented by Ar Brandon last updated on 05/Nov/20

��You'll have to get used-to it bro��

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