Question Number 108295 by bobhans last updated on 16/Aug/20 $$\:\:\:\:\frac{\mathbb{B}\mathrm{ob}\mathbb{H}\mathrm{ans}}{\beta\mathrm{o}\flat} \\ $$$$\:\left(\mathrm{1}\right)\begin{cases}{\left(\mathrm{x}+\mathrm{y}\right)\left(\mathrm{x}^{\mathrm{2}} −\mathrm{y}^{\mathrm{2}} \right)\:=\:\mathrm{9}}\\{\left(\mathrm{x}−\mathrm{y}\right)\left(\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} \right)\:=\:\mathrm{5}}\end{cases} \\ $$$$\mathrm{find}\:\mathrm{the}\:\mathrm{solution} \\ $$$$\left(\mathrm{2}\right)\:\mathrm{x}\:\frac{\mathrm{dy}}{\mathrm{dx}}\:=\:\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} \:\mathrm{when}\:\mathrm{x}=\mathrm{1}\:\mathrm{give}\:\mathrm{y}\:=\:\mathrm{2}\: \\ $$ Answered…
Question Number 173829 by mnjuly1970 last updated on 19/Jul/22 $$ \\ $$$$\:\:\:\:{If}\:\:,\:{x}\in\:\left[\mathrm{0}\:,\:\mathrm{1}\right]\: \\ $$$$\:\:\:\:\:\:,\:\:\:\:\mid\:\sqrt{\mathrm{1}−{x}^{\:\mathrm{2}} }\:−{ax}−{b}\:\mid\leqslant\:\frac{\sqrt{\mathrm{2}}\:−\mathrm{1}}{\mathrm{2}} \\ $$$$\:\:\:\:{find}\:{the}\:{values}\:{of}\:\:\left(\:{a}\:,\:{b}\:\right) \\ $$$$\:\:\:{a}\:,\:{b}\in\:\mathbb{R}. \\ $$$$ \\ $$ Commented by…
Question Number 173831 by dragan91 last updated on 19/Jul/22 $$\mathrm{Solve} \\ $$$$\mathrm{x}+\mathrm{y}=\mathrm{4} \\ $$$$\mathrm{x}^{\mathrm{x}} −\mathrm{y}^{\mathrm{y}} =\mathrm{13}\left(\mathrm{x}−\mathrm{y}\right) \\ $$ Answered by mr W last updated on…
Question Number 42756 by Aviral last updated on 02/Sep/18 $$\mathrm{33}\sqrt{\mathrm{67}} \\ $$ Commented by Joel578 last updated on 02/Sep/18 $$\approx\:\mathrm{270}.\mathrm{117} \\ $$ Terms of Service…
Question Number 173830 by dragan91 last updated on 19/Jul/22 $$ \\ $$$$\mathrm{How}\:\mathrm{many}\:\mathrm{positive}\:\mathrm{integer}\:\mathrm{multiples}\:\mathrm{of}\:\mathrm{2431} \\ $$$$\mathrm{can}\:\mathrm{be}\:\mathrm{expressed}\:\mathrm{in}\:\mathrm{the}\:\mathrm{form}\:\mathrm{7}^{\mathrm{a}} −\mathrm{7}^{\mathrm{b}} \\ $$$$\mathrm{where}\:\boldsymbol{{a}}\:\mathrm{and}\:\boldsymbol{{b}}\:\mathrm{are}\:\mathrm{integer}\:\mathrm{such}\:\mathrm{that}\:\mathrm{0}\leqslant\boldsymbol{{b}}\leqslant\boldsymbol{{a}}\leqslant\mathrm{2022}\: \\ $$ Terms of Service Privacy Policy Contact:…
Question Number 108291 by bemath last updated on 16/Aug/20 $$\:\:\:\frac{\parallel\:\mathcal{B}{e}\mathcal{M}{ath}\:\parallel}{°\int\:{dx}°} \\ $$$$\left(\mathrm{1}\right)\:{Given}\:\left({x}+\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }\right)\left({y}+\sqrt{\mathrm{1}+{y}^{\mathrm{2}} }\right)=\mathrm{1} \\ $$$${find}\:\left({x}+{y}\right)^{\mathrm{2}} \: \\ $$ Answered by john santu last updated…
Question Number 173820 by dragan91 last updated on 18/Jul/22 $$ \\ $$ Prove $$ \\ $$$$\sqrt[{\mathrm{3}}]{\mathrm{cos}\frac{\pi}{\mathrm{19}}\mathrm{cos}\frac{\mathrm{7}\pi}{\mathrm{19}}\mathrm{cos}\frac{\mathrm{8}\pi}{\mathrm{19}}}+\sqrt[{\mathrm{3}}]{\mathrm{cos}\frac{\mathrm{2}\pi}{\mathrm{19}}\mathrm{cos}\frac{\mathrm{3}\pi}{\mathrm{19}}\mathrm{cos}\frac{\mathrm{5}\pi}{\mathrm{19}}}− \\ $$$$−\sqrt[{\mathrm{3}}]{\mathrm{cos}\frac{\mathrm{4}\pi}{\mathrm{19}}\mathrm{cos}\frac{\mathrm{6}\pi}{\mathrm{19}}\mathrm{cos}\frac{\mathrm{9}\pi}{\mathrm{19}}}=\frac{\mathrm{1}}{\mathrm{2}}\sqrt[{\mathrm{3}}]{\mathrm{3}\sqrt[{\mathrm{3}}]{\mathrm{19}}−\mathrm{1}} \\ $$$$ \\ $$$$ \\ $$$$ \\…
Question Number 108270 by Muhsang S L last updated on 15/Aug/20 $$\mathrm{If}\:{x},\:{y},\:{z}\:>\:−\mathrm{1},\:\mathrm{show}\:\mathrm{that}\:\frac{\mathrm{1}\:+\:{x}^{\mathrm{2}} }{\mathrm{1}\:+\:{y}\:+\:{z}^{\mathrm{2}} }\:+\:\frac{\mathrm{1}\:+\:{y}^{\mathrm{2}} }{\mathrm{1}\:+\:{z}\:+\:{x}^{\mathrm{2}} }\:+\:\frac{\mathrm{1}\:+\:{z}^{\mathrm{2}} }{\mathrm{1}\:+\:{x}\:+\:{y}^{\mathrm{2}} }\:\geqslant\:\mathrm{2} \\ $$ Commented by udaythool last updated…
Question Number 173815 by mr W last updated on 18/Jul/22 $${prove}\: \\ $$$$\left(\frac{{n}+\mathrm{1}}{\mathrm{3}}\right)^{{n}} <{n}! \\ $$ Commented by mr W last updated on 19/Jul/22 $${good}\:{idea}\:{sir}!\:{thanks}!…
Question Number 173798 by payamnazerian last updated on 18/Jul/22 $$ \\ $$$$\:\:\:\:\:\:\:\left(\:\varphi^{\:\mathrm{5}} +\:\varphi^{\:\mathrm{4}} +\:\mathrm{1}\right)^{\:\mathrm{2}} =\:{m}\varphi\:+\:{n} \\ $$$$\:\:\:\:\:\:\:\:\:{m},{n}=? \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:…….. \\ $$$$\:\:\:\:\:\:\:\:\:\: \\ $$ Answered by…