Question Number 96505 by bemath last updated on 02/Jun/20 $$\frac{\sqrt{\sqrt[{\mathrm{4}\:\:}]{\mathrm{8}}−\sqrt{\sqrt{\mathrm{2}}+\mathrm{1}}}}{\:\sqrt{\sqrt[{\mathrm{4}\:\:}]{\mathrm{8}}+\sqrt{\sqrt{\mathrm{2}}−\mathrm{1}}}−\sqrt{\sqrt[{\mathrm{4}\:\:}]{\mathrm{8}}−\sqrt{\sqrt{\mathrm{2}}−\mathrm{1}}}}\:? \\ $$ Commented by bemath last updated on 02/Jun/20 $$\mathrm{yes}\:\mathrm{sir}.\:\mathrm{you}\:\mathrm{are}\:\mathrm{right} \\ $$ Commented by bobhans…
Question Number 30958 by rahul 19 last updated on 01/Mar/18 Commented by rahul 19 last updated on 01/Mar/18 Commented by rahul 19 last updated on…
Question Number 30956 by Tinkutara last updated on 01/Mar/18 Answered by MJS last updated on 01/Mar/18 $$\left(\mathrm{3}\right)\:\mathrm{because}\:{n}!\mid\underset{{k}={i}} {\overset{{i}+{n}−\mathrm{1}} {\prod}}{k};\:{i}\in\mathbb{N} \\ $$ Commented by Tinkutara last…
Question Number 162027 by mathlove last updated on 25/Dec/21 Answered by MJS_new last updated on 25/Dec/21 $${x}\in\mathbb{R}\wedge{y}\in\mathbb{R}\wedge{n}\in\mathbb{Z}:\:{y}=\sqrt[{\mathrm{2}{n}+\mathrm{1}}]{−{x}}=−\sqrt[{\mathrm{2}{n}+\mathrm{1}}]{{x}} \\ $$$$\Rightarrow \\ $$$$\sqrt[{\mathrm{3}}]{−\mathrm{9}^{−\mathrm{2}} }=−\sqrt[{\mathrm{3}}]{\mathrm{9}^{−\mathrm{2}} }=−\sqrt[{\mathrm{3}}]{\frac{\mathrm{1}}{\mathrm{9}^{\mathrm{2}} }}=−\frac{\mathrm{1}}{\:\sqrt[{\mathrm{3}}]{\mathrm{81}}}=−\frac{\mathrm{1}}{\mathrm{3}\sqrt[{\mathrm{3}}]{\mathrm{3}}} \\…
Question Number 162003 by mathlove last updated on 25/Dec/21 $$\sqrt[{\mathrm{4}}]{\mathrm{27}\sqrt[{\mathrm{4}}]{\mathrm{27}\sqrt[{\mathrm{4}}]{\mathrm{27}….\:}}}={x} \\ $$$$\sqrt{\mathrm{5}\sqrt{\mathrm{5}\sqrt{\mathrm{5}\sqrt{\mathrm{5}…..}}}}={y} \\ $$$${y}^{\mathrm{2}} −{x}^{\mathrm{2}} =? \\ $$ Answered by Rasheed.Sindhi last updated on 25/Dec/21…
Question Number 161999 by mahdipoor last updated on 25/Dec/21 $$\int\frac{{dx}}{\:\sqrt{{x}^{\mathrm{3}} −\mathrm{4}{x}}} \\ $$ Answered by aleks041103 last updated on 25/Dec/21 $$\int\frac{{dx}}{\:\sqrt{{x}^{\mathrm{3}} −\mathrm{4}{x}}}=\int\frac{{dx}}{\:\sqrt{{x}}}\:\frac{\mathrm{1}}{\:\sqrt{{x}^{\mathrm{2}} −\mathrm{4}}}= \\ $$$$=\mathrm{2}\int\frac{{d}\left(\sqrt{{x}}\right)}{\:\sqrt{\left(\sqrt{{x}}\right)^{\mathrm{4}}…
Question Number 96460 by bobhans last updated on 01/Jun/20 Commented by MJS last updated on 01/Jun/20 $$−\mathrm{4} \\ $$$$\mathrm{only}\:\mathrm{step}\:\mathrm{needed}\:\mathrm{several}\:\mathrm{times}\:\mathrm{is} \\ $$$$\frac{{a}}{{b}+\sqrt{{c}}}=\frac{{a}\left({b}−\sqrt{{c}}\right)}{{b}^{\mathrm{2}} −{c}} \\ $$ Commented…
Question Number 30916 by Tinkutara last updated on 28/Feb/18 Commented by Tinkutara last updated on 28/Feb/18 Answer is 1⃣ Answered by $@ty@m last updated on 28/Feb/18 $$\mathrm{ln}\:\left(\mathrm{1}+{x}\right)={x}−\frac{{x}^{\mathrm{2}}…
Question Number 161964 by HongKing last updated on 24/Dec/21 $$\mathrm{Prove}\:\mathrm{that}:\:\left(\mathrm{a}\:\mathrm{series}\:\mathrm{inspired}\:\mathrm{Knopp}\:\mathrm{Konrad}\right) \\ $$$$\sqrt{\mathrm{e}^{\boldsymbol{\pi}} }\:\:=\:\underset{\boldsymbol{\mathrm{k}}=\mathrm{0}} {\overset{\infty} {\sum}}\:\frac{\mathrm{sin}\left(\frac{{k}\pi}{\mathrm{4}}\right)}{\left(\mathrm{k}!\right)\:\sqrt{\mathrm{2}^{\boldsymbol{\mathrm{k}}} }}\:\:\pi^{\boldsymbol{\mathrm{k}}} \\ $$ Answered by mindispower last updated on 25/Dec/21…
Question Number 161952 by mathlove last updated on 24/Dec/21 $$!!\mathrm{8}=? \\ $$ Commented by Rasheed.Sindhi last updated on 24/Dec/21 $${You}\:{don}'{t}\:{see}\:{the}\:{answers}\:{of}\:{your} \\ $$$${questions},{I}\:{think}. \\ $$$${Only}\:{you}\:{put}\:{your}\:{questions}\:{on}\:{the} \\…