Question Number 162068 by HongKing last updated on 25/Dec/21 Answered by Lordose last updated on 26/Dec/21 $$ \\ $$$$\Omega\:=\:\underset{\epsilon\rightarrow\mathrm{0}} {\mathrm{lim}}\int_{\mathrm{sin}^{−\mathrm{1}} \left(\epsilon\right)} ^{\:\mathrm{sin}^{−\mathrm{1}} \left(\mathrm{1}−\epsilon\right)} \mathrm{log}\left(\left(\mathrm{cos}\left(\mathrm{x}\right)\right)^{\mathrm{cot}\left(\mathrm{x}\right)} \centerdot\left(\mathrm{sin}\left(\mathrm{x}\right)\right)^{\frac{\mathrm{cos}\left(\mathrm{x}\right)}{\mathrm{1}+\mathrm{sin}\left(\mathrm{x}\right)}}…
Question Number 30990 by rahul 19 last updated on 01/Mar/18 Answered by MJS last updated on 01/Mar/18 $$−\mathrm{2}{x}^{\mathrm{2}} +{kx}+\left({k}^{\mathrm{2}} +\mathrm{5}\right)=\mathrm{0} \\ $$$${x}_{\mathrm{1}} =\frac{{k}}{\mathrm{4}}−\frac{\sqrt{\mathrm{9}{k}^{\mathrm{2}} +\mathrm{40}}}{\mathrm{4}} \\…
Question Number 96527 by student work last updated on 02/Jun/20 $$\mathrm{proof}\:\mathrm{that}\:\mathrm{1}^{\mathrm{2}} +\mathrm{2}^{\mathrm{2}} +\mathrm{3}^{\mathrm{2}} +….+\mathrm{n}^{\mathrm{2}} =\frac{\mathrm{n}\left(\mathrm{2n}+\mathrm{1}\right)\left(\mathrm{n}+\mathrm{1}\right)}{\mathrm{6}} \\ $$ Commented by 06122004 last updated on 02/Jun/20 Answered…
Question Number 162062 by HongKing last updated on 25/Dec/21 $$\Omega\left(\alpha;\beta\right)\:=\underset{\:-\mathrm{1}} {\overset{\:\mathrm{1}} {\int}}\:\frac{\left(\mathrm{1}+\mathrm{x}\right)^{\mathrm{2}\boldsymbol{\alpha}-\mathrm{1}} \:\left(\mathrm{1}-\mathrm{x}\right)^{\mathrm{2}\boldsymbol{\beta}-\mathrm{1}} }{\left(\mathrm{1}+\mathrm{x}^{\mathrm{2}} \right)^{\boldsymbol{\alpha}+\boldsymbol{\beta}} }\:\mathrm{dx}\:;\:\alpha;\beta>\mathrm{0} \\ $$$$\mathrm{find}\:\mathrm{a}\:\mathrm{closed}\:\mathrm{form}\:\mathrm{and}\:\mathrm{prove}\:\mathrm{that}: \\ $$$$\Omega\left(\mathrm{3},\mathrm{5}\right)\:>\:\sqrt{\Omega\left(\mathrm{4},\mathrm{5}\right)\centerdot\Omega\left(\mathrm{3},\mathrm{6}\right)} \\ $$ Answered by mindispower…
Question Number 162042 by HongKing last updated on 25/Dec/21 $$\mathrm{let}\:\:\mathrm{a};\mathrm{b};\mathrm{c}\in\mathbb{R}\:\:\mathrm{such}\:\mathrm{that}\:\:\mathrm{a}+\mathrm{b}+\mathrm{c}=\mathrm{3} \\ $$$$\mathrm{prove}\:\mathrm{that}: \\ $$$$\mathrm{a}^{\mathrm{3}} \:+\:\mathrm{b}^{\mathrm{3}} \:+\:\mathrm{c}^{\mathrm{3}} \:\geqslant\:\mathrm{a}^{\mathrm{3}} \mathrm{b}\:+\:\mathrm{b}^{\mathrm{3}} \mathrm{c}\:+\:\mathrm{c}^{\mathrm{3}} \mathrm{a} \\ $$ Terms of Service…
Question Number 162043 by HongKing last updated on 25/Dec/21 $$\mathrm{Find}\:\mathrm{valu}\:\mathrm{of}\:\:\boldsymbol{\mathrm{x}}\:\:\mathrm{if}\:\:\mathrm{x}\in\mathbb{R}\: \\ $$$$\sqrt[{\mathrm{3}}]{\mathrm{9x}\:-\:\mathrm{1}}\:+\:\sqrt{\mathrm{8x}\:-\:\mathrm{1}}\:+\:\sqrt[{\mathrm{4}}]{\mathrm{8x}\:+\:\mathrm{15}}\:-\:\frac{\mathrm{5}}{\mathrm{2}}\:=\:\mathrm{0} \\ $$ Commented by mr W last updated on 25/Dec/21 $${with}\:{x}=\frac{\mathrm{1}}{\mathrm{8}} \\ $$$$\sqrt[{\mathrm{3}}]{\frac{\mathrm{9}}{\mathrm{8}}−\mathrm{1}}+\sqrt{\frac{\mathrm{8}}{\mathrm{8}}−\mathrm{1}}+\sqrt[{\mathrm{4}}]{\frac{\mathrm{8}}{\mathrm{8}}+\mathrm{15}}−\frac{\mathrm{5}}{\mathrm{2}}…
Question Number 96505 by bemath last updated on 02/Jun/20 $$\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}}} \\…