Question Number 122377 by aristarque last updated on 16/Nov/20 Commented by Dwaipayan Shikari last updated on 16/Nov/20 $$\sqrt[{\mathrm{3}}]{\mathrm{54}\sqrt{\mathrm{3}}+\mathrm{41}\sqrt{\mathrm{5}}}\:=\sqrt[{\mathrm{3}}]{\left(\mathrm{2}\sqrt{\mathrm{3}}+\sqrt{\mathrm{5}}\right)^{\mathrm{3}} } \\ $$$$\sqrt[{\mathrm{3}}]{\mathrm{54}\sqrt{\mathrm{3}}−\mathrm{41}\sqrt{\mathrm{5}}}=\sqrt{\left(\mathrm{2}\sqrt{\mathrm{3}}−\sqrt{\mathrm{5}}\right)^{\mathrm{3}} } \\ $$$${A}=\frac{\mathrm{2}\sqrt{\mathrm{3}}+\sqrt{\mathrm{5}}+\mathrm{2}\sqrt{\mathrm{3}}−\sqrt{\mathrm{5}}}{\:\sqrt{\mathrm{3}}}=\mathrm{4} \\…
Question Number 56823 by tanmay.chaudhury50@gmail.com last updated on 24/Mar/19 Commented by tanmay.chaudhury50@gmail.com last updated on 25/Mar/19 $${Bengali}\:+\left({broken}\:{hindi}+{broken}\:{english}\right)=\mathrm{2} \\ $$ Answered by peter frank last updated…
Question Number 122336 by aristarque last updated on 15/Nov/20 $${please}\:{prove}\:{that}\:{sup}\left(−{A}\right)=−{inf}\left({A}\right) \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
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Question Number 122322 by EngLewis last updated on 15/Nov/20 $$\mid{x}+\mathrm{1}\mid<\mid{x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{2}\mid \\ $$$${solve}\:{for}\:{x} \\ $$ Answered by mathmax by abdo last updated on 15/Nov/20 $$\mathrm{we}\:\mathrm{have}\:\mathrm{x}^{\mathrm{2}}…
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Question Number 122317 by Dwaipayan Shikari last updated on 15/Nov/20 $$\mathrm{1}+\mathrm{9}\left(\frac{\mathrm{1}}{\mathrm{4}}\right)^{\mathrm{4}} +\mathrm{17}\left(\frac{\mathrm{1}.\mathrm{5}}{\mathrm{4}.\mathrm{8}}\right)^{\mathrm{4}} +\mathrm{25}\left(\frac{\mathrm{1}.\mathrm{5}.\mathrm{9}}{\mathrm{4}.\mathrm{8}.\mathrm{12}}\right)^{\mathrm{4}} +…=\frac{\pi}{\mathrm{2}\sqrt{\mathrm{2}}\left(\Gamma^{\mathrm{2}} \left(\frac{\mathrm{3}}{\mathrm{4}}\right)\right)} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 56775 by zambolly19 last updated on 23/Mar/19 $${if}\:{x},{y}\backepsilon\Re,{show}\:{that}\:\mid{x}+{y}\mid=\mid{x}\mid+\mid{y}\mid\:{iff}\:{xy}\geqslant\mathrm{0} \\ $$ Commented by maxmathsup by imad last updated on 23/Mar/19 $${we}\:{have}\:\mid{x}+{y}\mid^{\mathrm{2}} −\left(\mid{x}\mid+\mid{y}\mid\right)^{\mathrm{2}} \:={x}^{\mathrm{2}} +\mathrm{2}{xy}+{y}^{\mathrm{2}}…
Question Number 122302 by Dwaipayan Shikari last updated on 15/Nov/20 $$\int_{\mathrm{0}} ^{\infty} \frac{{dx}}{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)\left(\mathrm{4}+{x}^{\mathrm{2}} \right)\left(\mathrm{16}+{x}^{\mathrm{2}} \right)\left(\mathrm{64}+{x}^{\mathrm{2}} \right)} \\ $$ Answered by TANMAY PANACEA last updated…
Question Number 122303 by Dwaipayan Shikari last updated on 15/Nov/20 $$\frac{{log}\mathrm{1}}{\:\sqrt{\mathrm{1}}}−\frac{{log}\mathrm{3}}{\:\sqrt{\mathrm{3}}}+\frac{{log}\mathrm{5}}{\:\sqrt{\mathrm{5}}}−\frac{{log}\mathrm{7}}{\:\sqrt{\mathrm{7}}}+..{C}=\left(\frac{\pi}{\mathrm{4}}−\frac{\gamma}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{2}}{log}\left(\mathrm{2}\pi\right)\right)\left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}}}−\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}+..{C}\right) \\ $$$$\gamma\:=\mathscr{E}{ulerian}\:{constant} \\ $$ Commented by rs4089 last updated on 15/Nov/20 $${i}\:{think}\:{its}\:{expansion}\:{is}\:{given}\:{by}\:{indian}\:{mathematician}\:\:{srinivasa}\:{ramanujan}\:{to}\:{profecer}\:{g}.{h}.\:{hardy}\:. \\ $$…