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 $${You}\:{don}'{t}\:{see}\:{the}\:{answers}\:{of}\:{your} \\ $$$${questions},{I}\:{think}. \\ $$$${Only}\:{you}\:{put}\:{your}\:{questions}\:{on}\:{the} \\…
Question Number 161951 by mathlove last updated on 24/Dec/21 $${x}^{{x}} =\mathrm{2}^{\mathrm{2048}} \\ $$$${x}=? \\ $$ Answered by aleks041103 last updated on 24/Dec/21 $$\mathrm{2048}=\mathrm{2}^{\mathrm{11}} \\ $$$$\Rightarrow{x}^{{x}}…
Question Number 161946 by mathlove last updated on 24/Dec/21 Answered by Rasheed.Sindhi last updated on 29/Dec/21 $${x}+{y}\left({x}+\mathrm{1}\right)=\mathrm{3}\Rightarrow{y}=\frac{\mathrm{3}−{x}}{{x}+\mathrm{1}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\Rightarrow{y}+\mathrm{1}=\frac{\mathrm{3}−{x}+{x}+\mathrm{1}}{{x}+\mathrm{1}}=\frac{\mathrm{4}}{{x}+\mathrm{1}} \\ $$$${y}+{z}\left({y}+\mathrm{1}\right)=\mathrm{5}\Rightarrow\frac{\mathrm{3}−{x}}{{x}+\mathrm{1}}+{z}\left(\frac{\mathrm{4}}{{x}+\mathrm{1}}\right)=\mathrm{5} \\ $$$$\:\:\:\:\:\:\:\Rightarrow\mathrm{3}−{x}+\mathrm{4}{z}=\mathrm{5}{x}+\mathrm{5}\Rightarrow{x}=\frac{\mathrm{2}{z}−\mathrm{1}}{\mathrm{3}} \\ $$$${z}+{x}\left({z}+\mathrm{1}\right)=\mathrm{7}\Rightarrow{z}+\left(\frac{\mathrm{2}{z}−\mathrm{1}}{\mathrm{3}}\right)\left({z}+\mathrm{1}\right)=\mathrm{7}…
Question Number 96407 by Enyz last updated on 01/Jun/20 Commented by prakash jain last updated on 01/Jun/20 $${z}^{\mathrm{3}} +\mathrm{2}{z}^{\mathrm{2}} +\left({k}−\mathrm{8}\sqrt{\mathrm{2}}{i}\right){z}+\left(\mathrm{8}−\mathrm{4}{k}\sqrt{\mathrm{2}}{i}\right)=\mathrm{0} \\ $$$$\mathrm{Let}\:{z}={x}\:\mathrm{be}\:\mathrm{real}\:\mathrm{root} \\ $$$${x}^{\mathrm{3}} +\mathrm{2}{x}^{\mathrm{2}}…
Question Number 30860 by ajfour last updated on 27/Feb/18 $${S}=\:\mathrm{3}\left(\mathrm{1}!\right)−\mathrm{4}\left(\mathrm{2}!\right)+\mathrm{5}\left(\mathrm{3}!\right)−\mathrm{6}\left(\mathrm{4}!\right)+…. \\ $$$$\:\:\:\:…..−\left(\mathrm{2008}\right)\left(\mathrm{2006}!\right)+\mathrm{2007}! \\ $$$${Find}\:{value}\:{of}\:{S}. \\ $$ Answered by MJS last updated on 27/Feb/18 $$\mathrm{3}\centerdot\mathrm{1}!−\mathrm{4}\centerdot\mathrm{2}!=−\mathrm{5}\:\left[=−\left(\mathrm{3}!−\mathrm{1}\right)\right] \\…