Question Number 159066 by amin96 last updated on 12/Nov/21 $$\mathrm{16}/\mathrm{4}\left(\mathrm{2}+\mathrm{2}\right)=? \\ $$ Commented by Rasheed.Sindhi last updated on 12/Nov/21 $$\mathrm{2}\:\mathrm{Conventions}: \\ $$$$\left(\mathrm{i}\right)\mathrm{implied}\:\mathrm{multiplication}\:\mathrm{has}\:\mathrm{priority} \\ $$$$\:\:\:\:\:\:\:\mathrm{over}\:\mathrm{division}: \\…
Question Number 27996 by JI Siam last updated on 18/Jan/18 $$\mathrm{p},\mathrm{q}\:\mathrm{are}\:\mathrm{two}\:\mathrm{natural}\:\mathrm{number}\:\mathrm{and}\: \\ $$$$\:\frac{\mathrm{p}^{\mathrm{6}} +\mathrm{2p}^{\mathrm{4}} +\mathrm{4p}^{\mathrm{2}} }{\mathrm{p}^{\mathrm{9}} −\mathrm{8p}^{\mathrm{3}} }−\frac{\mathrm{1}}{\mathrm{4q}}=\frac{\mathrm{5}}{\mathrm{6q}}, \\ $$$$\mathrm{then}\:\mathrm{find}\:\mathrm{the}\:\mathrm{minimum} \\ $$$$\mathrm{possible}\:\mathrm{value}\:\mathrm{of}\:\mathrm{p}+\mathrm{q} \\ $$ Answered…
Question Number 159057 by HongKing last updated on 12/Nov/21 $$\mathrm{Solve}\:\mathrm{for}\:\mathrm{positive}\:\mathrm{integers}: \\ $$$$\mathrm{a}^{\mathrm{3}} \:+\:\mathrm{9b}^{\mathrm{2}} \:+\:\mathrm{9c}^{\mathrm{2}} \:=\:\mathrm{2017} \\ $$$$\mathrm{where}\:\:\mathrm{a}\:\geqslant\:\mathrm{b}\:\geqslant\:\mathrm{c} \\ $$$$ \\ $$ Commented by Rasheed.Sindhi last…
Question Number 159049 by mnjuly1970 last updated on 12/Nov/21 $$ \\ $$$$\:\:\:\:\:\:{solve}\:{the}\:{following}\:{equation}: \\ $$$$\: \\ $$$${sin}^{\:\mathrm{6}} \left({x}\right)\:+{cos}^{\:\mathrm{6}} \left({x}\right)+{sin}^{\mathrm{4}} \left({x}\right)+{cos}^{\:\mathrm{4}} \left({x}\right)=\frac{\mathrm{3}}{\mathrm{4}} \\ $$$$ \\ $$ Answered…
Question Number 27983 by JI Siam last updated on 18/Jan/18 $$\left.\mathrm{1}\right)\:\mathrm{find}\:\mathrm{two}\:\:\mathrm{factors}\:\mathrm{of}\:\mathrm{1000001}\:\mathrm{other}\:\mathrm{than}\:\mathrm{1}\:\mathrm{and}\:\mathrm{1000001} \\ $$$$\left.\mathrm{2}\right)\left(\mathrm{x}^{\mathrm{2}} −\mathrm{5x}+\mathrm{5}\right)^{\left(\mathrm{x}^{\mathrm{2}} +\mathrm{2x}−\mathrm{24}\right)} =\mathrm{1}\:\mathrm{what}\:\mathrm{is}\:\mathrm{the}\:\mathrm{value}\:\:\mathrm{of}\:\mathrm{the}\:\mathrm{product}\:\mathrm{of}\:\mathrm{the}\:\mathrm{solutions}? \\ $$ Answered by mrW2 last updated on 18/Jan/18…
Question Number 27977 by NECx last updated on 18/Jan/18 $$\mid\mathrm{2}{x}+\mathrm{1}\mid\leqslant\mathrm{2} \\ $$ Answered by Rasheed.Sindhi last updated on 18/Jan/18 $$\mid\mathrm{2x}+\mathrm{1}\mid\leqslant\mathrm{2} \\ $$$$\pm\left(\mathrm{2x}+\mathrm{1}\right)\leqslant\mathrm{2} \\ $$$$\mathrm{2x}+\mathrm{1}\leqslant\mathrm{2}\:\wedge\:−\mathrm{2x}−\mathrm{1}\leqslant\mathrm{2} \\…
Question Number 159045 by amin96 last updated on 12/Nov/21 $$\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\left(\mathrm{3}{n}+\mathrm{1}\right)^{\mathrm{3}} }=? \\ $$ Answered by mindispower last updated on 16/Nov/21 $$=\underset{{n}\geqslant\mathrm{0}} {\sum}.\frac{\mathrm{1}}{\mathrm{27}}\frac{\mathrm{1}}{\left({n}+\frac{\mathrm{1}}{\mathrm{3}}\right)^{\mathrm{3}} }…
Question Number 27976 by NECx last updated on 18/Jan/18 $${solve} \\ $$$$ \\ $$$$\frac{\mathrm{2}{x}}{{x}^{\mathrm{2}} +\mathrm{1}}<\frac{\mathrm{3}{x}+\mathrm{1}}{\mathrm{2}\left({x}^{\mathrm{2}} +\mathrm{1}\right)} \\ $$$$ \\ $$ Commented by abdo imad last…
Question Number 27975 by NECx last updated on 18/Jan/18 $${solve}\:{the}\:{inequality} \\ $$$$\frac{\mathrm{1}}{{x}^{\mathrm{2}} +{x}+\mathrm{1}}>\mathrm{0} \\ $$ Answered by mrW2 last updated on 18/Jan/18 $$\frac{\mathrm{1}}{{x}^{\mathrm{2}} +{x}+\mathrm{1}}=\frac{\mathrm{1}}{{x}^{\mathrm{2}} +\mathrm{2}×\frac{\mathrm{1}}{\mathrm{2}}{x}+\left(\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}}…
Question Number 159046 by mnjuly1970 last updated on 12/Nov/21 Commented by cortano last updated on 13/Nov/21 $${f}\left({x}\right)=\frac{\mathrm{3}\left(\mathrm{1}−\mathrm{cos}\:^{\mathrm{2}} {x}\right)^{\mathrm{2}} +\mathrm{2}}{\mathrm{6cos}\:^{\mathrm{2}} {x}+\mathrm{1}} \\ $$$${f}\left({x}\right)=\frac{\mathrm{3}\left(\mathrm{cos}\:^{\mathrm{4}} {x}−\mathrm{2cos}\:^{\mathrm{2}} {x}+\mathrm{1}\right)+\mathrm{2}}{\mathrm{6cos}\:^{\mathrm{2}} {x}+\mathrm{1}}…