Question Number 176636 by cortano1 last updated on 23/Sep/22 $$\:\begin{cases}{{p}^{\mathrm{3}} +{q}^{\mathrm{3}} ={r}^{\mathrm{2}} }\\{{p}^{\mathrm{3}} +{r}^{\mathrm{3}} ={q}^{\mathrm{2}} }\\{{q}^{\mathrm{3}} +{r}^{\mathrm{3}} ={p}^{\mathrm{2}} }\end{cases} \\ $$$$\:\Rightarrow\mathrm{20}{pqr}\:=? \\ $$ Commented by…
Question Number 45546 by Tawa1 last updated on 14/Oct/18 $$\mathrm{Simplify}:\:\:\:\:\:\sqrt[{\mathrm{3}}]{\sqrt{\mathrm{5}}\:\:+\:\:\mathrm{2}}\:\:\:+\:\:\:\:\sqrt[{\mathrm{3}}]{\sqrt{\mathrm{5}}\:\:−\:\mathrm{2}}\:\: \\ $$ Commented by Meritguide1234 last updated on 14/Oct/18 $${x}^{\mathrm{3}} =\left(\sqrt{\mathrm{5}}+\mathrm{2}\right)+\left(\sqrt{\mathrm{5}}−\mathrm{2}\right)+\mathrm{3}.{x} \\ $$$$ \\ $$…
Question Number 111062 by malwan last updated on 01/Sep/20 Commented by malwan last updated on 01/Sep/20 $${please} \\ $$$${I}\:{need}\:{all}\:{solutions}\:{for}\:{exam}\: \\ $$ Commented by malwan last…
Question Number 176598 by Rasheed.Sindhi last updated on 22/Sep/22 $${x}^{\mathrm{3}} +\frac{\mathrm{1}}{{x}^{\mathrm{3}} }=\mathrm{1} \\ $$$$\frac{\left({x}^{\mathrm{5}} +\frac{\mathrm{1}}{{x}^{\mathrm{5}} }\right)^{\mathrm{3}} −\mathrm{1}}{{x}^{\mathrm{5}} +\frac{\mathrm{1}}{{x}^{\mathrm{5}} }}=? \\ $$$${Q}#\mathrm{176387}\:{reposted}\:{for}\:{a}\:{new}\:{answer}. \\ $$ Answered by…
Question Number 176592 by Shrinava last updated on 22/Sep/22 $$\mathrm{Solve}\:\mathrm{the}\:\mathrm{equation}: \\ $$$$\mathrm{2}^{\boldsymbol{\mathrm{x}}} \:+\:\mathrm{3}^{\boldsymbol{\mathrm{x}}} \:−\:\mathrm{4}^{\boldsymbol{\mathrm{x}}} \:+\:\mathrm{6}^{\boldsymbol{\mathrm{x}}} \:−\:\mathrm{9}^{\boldsymbol{\mathrm{x}}} \:=\:\mathrm{1} \\ $$ Answered by Frix last updated on…
Question Number 45512 by Tawa1 last updated on 13/Oct/18 $$\mathrm{Calculate}:\:\:\:\frac{\left(\mathrm{2}^{\mathrm{4}} \:+\:\frac{\mathrm{1}}{\mathrm{4}}\right)\:\left(\mathrm{4}^{\mathrm{4}} \:+\:\frac{\mathrm{1}}{\mathrm{4}}\right)\left(\mathrm{6}^{\mathrm{4}} \:+\:\frac{\mathrm{1}}{\mathrm{4}}\right)\left(\mathrm{8}^{\mathrm{4}} \:+\:\frac{\mathrm{1}}{\mathrm{4}}\right)\left(\mathrm{10}^{\mathrm{4}} \:+\:\frac{\mathrm{1}}{\mathrm{4}}\right)\left(\mathrm{12}^{\mathrm{4}} \:+\:\frac{\mathrm{1}}{\mathrm{4}}\right)}{\left(\mathrm{1}^{\mathrm{4}} \:+\:\frac{\mathrm{1}}{\mathrm{4}}\right)\left(\mathrm{3}^{\mathrm{4}} \:+\:\frac{\mathrm{1}}{\mathrm{4}}\right)\:\left(\mathrm{5}^{\mathrm{4}} \:+\:\frac{\mathrm{1}}{\mathrm{4}}\right)\:\left(\mathrm{7}^{\mathrm{4}} \:+\:\frac{\mathrm{1}}{\mathrm{4}}\right)\:\left(\mathrm{9}^{\mathrm{4}} \:+\:\frac{\mathrm{1}}{\mathrm{4}}\right)\left(\mathrm{11}^{\mathrm{4}} \:+\:\frac{\mathrm{1}}{\mathrm{4}}\right)} \\ $$ Commented…
Question Number 176580 by mathlove last updated on 22/Sep/22 $$\mathrm{4}^{{x}^{\mathrm{2}} −\mathrm{2}{x}+\mathrm{2}} −\mathrm{2}^{{x}^{\mathrm{2}} −\mathrm{2}{x}+\mathrm{3}} +\mathrm{2}=\mathrm{2}^{{x}^{\mathrm{2}} −\mathrm{2}{x}+\mathrm{2}} \\ $$$${x}=? \\ $$ Answered by cortano1 last updated on…
Question Number 111029 by Khanacademy last updated on 01/Sep/20 Answered by mindispower last updated on 02/Sep/20 $$=\int_{\mathrm{0}} ^{\pi} \frac{\left(\pi−{x}\right)^{\mathrm{2}} {sin}\left(\mathrm{2}\pi−\mathrm{2}{x}\right){sin}\left(\frac{\pi}{\mathrm{2}}{cos}\left(\pi−{x}\right)\right)}{\pi−\mathrm{2}{x}}{dx} \\ $$$$=−\int_{\mathrm{0}} ^{\pi} \frac{\left(\pi^{\mathrm{2}} −\mathrm{2}{x}\pi+{x}^{\mathrm{2}}…
Question Number 176554 by mnjuly1970 last updated on 21/Sep/22 $$ \\ $$$$\:\:\:\:{x}\:+\:\frac{\mathrm{1}}{{x}}\:\:=\varphi\:\:\left({Golden}\:{ratio}\right) \\ $$$$\:\:\:{then} \\ $$$$\:\:\:\:\:\:\:\:\:{x}^{\:\mathrm{2000}} \:+\frac{\:\mathrm{1}}{{x}^{\:\mathrm{2000}} }\:=\:? \\ $$$$ \\ $$ Answered by Peace…
Question Number 176555 by Shrinava last updated on 21/Sep/22 $$\mathrm{ABCD}−\mathrm{convex}\:\mathrm{quadrilateral} \\ $$$$\mathrm{M}\in\mathrm{Int}\left(\mathrm{ABCD}\right)\:,\:\mathrm{F}−\mathrm{area}\:,\:\mathrm{s}−\mathrm{semiperimetr} \\ $$$$\mathrm{a}\:,\:\mathrm{b}\:,\:\mathrm{c}−\mathrm{sides}.\:\mathrm{Prove}\:\mathrm{that}: \\ $$$$\frac{\mathrm{MA}^{\mathrm{4}} }{\mathrm{b}}\:+\:\frac{\mathrm{MB}^{\mathrm{4}} }{\mathrm{c}^{\mathrm{4}} }\:+\:\frac{\mathrm{MC}^{\mathrm{4}} }{\mathrm{d}}\:+\:\frac{\mathrm{MD}^{\mathrm{4}} }{\mathrm{a}}\:\geqslant\:\frac{\mathrm{2F}^{\mathrm{2}} }{\mathrm{s}} \\ $$ Commented…