Question Number 50084 by ggny last updated on 13/Dec/18 $$\mathrm{could}\:\mathrm{you}\:\mathrm{help}\:\mathrm{me}\:\mathrm{sir}? \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 50083 by ggny last updated on 13/Dec/18 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 50064 by mondodotto@gmail.com last updated on 13/Dec/18 $$\boldsymbol{\mathrm{if}}\:\boldsymbol{\mathrm{y}}^{\boldsymbol{\mathrm{x}}} =\boldsymbol{\mathrm{x}}^{\boldsymbol{\mathrm{y}}} \: \\ $$$$\boldsymbol{\mathrm{and}}\:\boldsymbol{\mathrm{y}}\:\boldsymbol{\mathrm{and}}\:\boldsymbol{\mathrm{x}}\:\boldsymbol{\mathrm{are}}\:\boldsymbol{\mathrm{both}}\:\boldsymbol{\mathrm{not}} \\ $$$$\boldsymbol{\mathrm{equal}}\:\boldsymbol{\mathrm{to}}\:\mathrm{0}\:\boldsymbol{\mathrm{and}}\:\boldsymbol{\mathrm{l}} \\ $$$$\boldsymbol{\mathrm{find}}\:\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{value}}\:\boldsymbol{\mathrm{of}}\:\boldsymbol{\mathrm{x}}\:\boldsymbol{\mathrm{and}}\:\boldsymbol{\mathrm{y}} \\ $$ Answered by Necxx last updated…
Question Number 115597 by aurpeyz last updated on 26/Sep/20 Answered by 1549442205PVT last updated on 27/Sep/20 $$\left.\mathrm{4a}\right)\mathrm{Solve}\:\mathrm{the}\:\mathrm{equation}: \\ $$$$\frac{\mathrm{1}}{\:\sqrt{\mathrm{x}+\mathrm{1}}−\sqrt{\mathrm{x}}}−\frac{\mathrm{2}}{\:\sqrt{\mathrm{x}−\mathrm{2}}+\sqrt{\mathrm{x}}}=\sqrt{\mathrm{x}−\mathrm{1}} \\ $$$$\Leftrightarrow\frac{\sqrt{\mathrm{x}+\mathrm{1}}+\sqrt{\mathrm{x}}}{\mathrm{1}}−\frac{\mathrm{2}\left(\sqrt{\mathrm{x}}−\sqrt{\mathrm{x}−\mathrm{2}}\right)}{\mathrm{2}}=\sqrt{\mathrm{x}−\mathrm{1}} \\ $$$$\Leftrightarrow\sqrt{\mathrm{x}+\mathrm{1}}+\sqrt{\mathrm{x}}−\sqrt{\mathrm{x}}+\sqrt{\mathrm{x}−\mathrm{2}}=\sqrt{\mathrm{x}−\mathrm{1}} \\ $$$$\Leftrightarrow\sqrt{\mathrm{x}+\mathrm{1}}+\sqrt{\mathrm{x}−\mathrm{2}}=\sqrt{\mathrm{x}−\mathrm{1}}\:\left(\mathrm{1}\right)…
Question Number 115595 by floor(10²Eta[1]) last updated on 26/Sep/20 $$\mathrm{Prove}\:\mathrm{that},\:\mathrm{for}\:\mathrm{all}\:\mathrm{primes}\:\mathrm{p}>\mathrm{3}, \\ $$$$\mathrm{13}\mid\mathrm{10}^{\mathrm{2p}} −\mathrm{10}^{\mathrm{p}} +\mathrm{1} \\ $$ Answered by MJS_new last updated on 27/Sep/20 $$\mathrm{10}^{\mathrm{6}{n}} \equiv\mathrm{1mod13}…
Question Number 115575 by Aziztisffola last updated on 26/Sep/20 $$\underset{\mathrm{n}\rightarrow\infty} {\mathrm{lim}}\:\underset{\mathrm{k}=\mathrm{1}} {\overset{\mathrm{n}} {\prod}}\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{k}+\mathrm{1}}\right)=? \\ $$ Commented by Dwaipayan Shikari last updated on 26/Sep/20 $$\mathrm{First}\:\mathrm{term}\:\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{1}}\right)=\mathrm{0} \\…
Question Number 50017 by ajfour last updated on 13/Dec/18 $$\mathcal{CONGRATULATIONS}\:! \\ $$$${Tinkutara}-\:{our}\:{forum}\:{on} \\ $$$${having}\:{above}\:\mathrm{50000}\:{question}\:{posts}. \\ $$ Commented by rahul 19 last updated on 13/Dec/18 $${Wow}!…
Question Number 115555 by ZiYangLee last updated on 26/Sep/20 $$\mathrm{Given}\:\mathrm{that}\:{x},{y}\in\mathbb{R}\:\forall\:{x}^{\mathrm{2}} −{y}^{\mathrm{2}} =\mathrm{32}, \\ $$$$\left({x}+{y}\right)^{\mathrm{4}} +\left({x}−{y}\right)^{\mathrm{4}} =\mathrm{4352},\: \\ $$$$\mathrm{Find}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:{x}^{\mathrm{2}} +{y}^{\mathrm{2}} . \\ $$ Commented by Rasheed.Sindhi…
Question Number 181084 by SANOGO last updated on 21/Nov/22 Answered by mr W last updated on 21/Nov/22 $${A}=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\left(\mathrm{2}{n}−\mathrm{1}\right)^{\mathrm{2}} } \\ $$$${B}=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\left(\mathrm{2}{n}\right)^{\mathrm{2}}…
Question Number 50011 by ggny last updated on 13/Dec/18 Answered by tanmay.chaudhury50@gmail.com last updated on 13/Dec/18 $$\left.\mathrm{1}\right)\frac{\left(\mathrm{13}\right)^{\mathrm{4}} ×\left(\mathrm{3}×\mathrm{2}\right)^{\mathrm{3}} ×\mathrm{2}^{\mathrm{4}} ×\left(\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{6}} }{\mathrm{3}^{\mathrm{3}} ×\mathrm{13}^{\mathrm{3}} } \\ $$$$=\frac{\mathrm{13}^{\mathrm{4}}…