Question Number 72093 by naka3546 last updated on 24/Oct/19 $${a},\:{b},\:{c},\:{d},\:{e},\:{f},\:{g},\:{h},\:{i},\:{j}\:\:\:{are}\:\:\:{different}\:\:{nonnegative}\:\:{integers}\:. \\ $$$${Find}\:\:{all}\:\:{numbers}\:\:{that}\:\:{satisfy} \\ $$$$\:\:\:\:\:{a}\:×\:{bc}\:×\:{def}\:\:=\:\:{ghij} \\ $$ Commented by prakash jain last updated on 24/Oct/19 $${abcdef}={ghij}?…
Question Number 72086 by wo1lxjwjdb last updated on 24/Oct/19 $$\left(\mathrm{sin}^{−\mathrm{1}} \left({x}\right)\right)^{{v}} +\left(\mathrm{cos}^{−\mathrm{1}} \left({x}\right)^{{v}} \right) \\ $$$${for}\:{any}\:{value}\:{v}\:{what}\:{is}\:{it}\: \\ $$$${in}\:{terms}\:{of}\:{pi} \\ $$$$ \\ $$ Terms of Service…
Question Number 137615 by mohammad17 last updated on 04/Apr/21 $$\int_{\mathrm{0}} ^{\:\frac{\pi}{\mathrm{2}}} {ln}\left(\frac{\mathrm{1}+{tanx}}{\mathrm{1}−{tanx}}\right)^{{xcos}\left(\mathrm{8}{x}\right)} {dx} \\ $$ Commented by mohammad17 last updated on 04/Apr/21 $${hwo}\:{is}\:{can}\:{solve}\:{this}\:? \\ $$…
Question Number 137594 by Ñï= last updated on 04/Apr/21 $$\left({x}+\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}\right)\left({y}+\sqrt{{y}^{\mathrm{4}} +\mathrm{4}}\right)=\mathrm{9} \\ $$$${x}\sqrt{{y}^{\mathrm{4}} +\mathrm{4}}+{y}\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}=? \\ $$ Answered by bemath last updated on 04/Apr/21…
Question Number 137591 by Ñï= last updated on 04/Apr/21 $${a}=\sqrt[{\mathrm{3}}]{\mathrm{4}}+\sqrt[{\mathrm{3}}]{\mathrm{2}}+\sqrt[{\mathrm{3}}]{\mathrm{1}} \\ $$$$\frac{\mathrm{3}}{{a}}+\frac{\mathrm{3}}{{a}^{\mathrm{2}} }+\frac{\mathrm{1}}{{a}^{\mathrm{3}} }=? \\ $$ Commented by mr W last updated on 05/Apr/21 $$=\mathrm{1}…
Question Number 137590 by Ñï= last updated on 04/Apr/21 $${x}=\mathrm{1}+\frac{\pi+\left(\pi+\mathrm{1}\right)^{\mathrm{2}} +\left(\pi+\mathrm{2}\right)^{\mathrm{3}} +\left(\pi+\mathrm{3}\right)^{\mathrm{4}} }{\mathrm{4}+\mathrm{5}^{\mathrm{2}} +\mathrm{6}^{\mathrm{3}} +\mathrm{7}^{\mathrm{4}} } \\ $$$$\sqrt{{x}+\mathrm{2}\sqrt{{x}−\mathrm{1}}}+\sqrt{{x}−\mathrm{2}\sqrt{{x}−\mathrm{1}}}=? \\ $$ Commented by mindispower last updated…
Question Number 72048 by ahmadshahhimat775@gmail.com last updated on 23/Oct/19 Answered by mind is power last updated on 23/Oct/19 $$\mathrm{a}_{\mathrm{2}} =\frac{\mathrm{6}}{\mathrm{1}+\mathrm{3}}=\frac{\mathrm{3}}{\mathrm{2}}=\frac{\mathrm{3}.\mathrm{2}^{\mathrm{0}} }{\mathrm{2}}=\frac{\mathrm{2}+\mathrm{1}}{\mathrm{2}}=\frac{\mathrm{2}+\mathrm{2}^{\mathrm{0}} }{\mathrm{2}} \\ $$$$\mathrm{a}_{\mathrm{3}} =\frac{\mathrm{3}}{\mathrm{1}+\frac{\mathrm{3}}{\mathrm{2}}}=\frac{\mathrm{6}}{\mathrm{5}}\:=\frac{\mathrm{3}.\mathrm{2}}{\mathrm{3}.\mathrm{2}−\mathrm{1}}=\frac{\mathrm{4}+\mathrm{2}}{\mathrm{4}+\mathrm{2}−\mathrm{1}}=\frac{\mathrm{2}+\mathrm{2}^{\mathrm{2}}…
Question Number 72046 by aliesam last updated on 23/Oct/19 Commented by mathmax by abdo last updated on 26/Oct/19 $${let}\:{U}_{{n}} =\int_{\mathrm{0}} ^{\mathrm{1}} {nln}\left(\mathrm{1}+\left(\frac{{x}}{{n}}\right)^{\alpha} \right){dx}\:\Rightarrow{U}_{{n}} =_{\frac{{x}}{{n}}={t}} \:\:\:\int_{\mathrm{0}}…
Question Number 72047 by aliesam last updated on 23/Oct/19 Commented by mind is power last updated on 23/Oct/19 $$\mathrm{x}_{\mathrm{1}} =\mathrm{1},\mathrm{x}_{\mathrm{2}} =\mathrm{4},\mathrm{x}_{\mathrm{3}} =\mathrm{9},\mathrm{x}_{\mathrm{4}} =\mathrm{16} \\ $$$$\forall_{\mathrm{n}}…
Question Number 137582 by SOMEDAVONG last updated on 04/Apr/21 $$\mathrm{A}=\sqrt[{\mathrm{3}}]{\frac{\mathrm{1}}{\mathrm{3}}\left(\sqrt[{\mathrm{3}}]{\mathrm{2}}−\mathrm{1}\right)\left(\sqrt[{\mathrm{3}}]{\mathrm{2}}+\mathrm{1}\right)^{\mathrm{3}} } \\ $$ Answered by Ñï= last updated on 04/Apr/21 $${A}=\sqrt[{\mathrm{3}}]{\frac{\mathrm{1}}{\mathrm{3}}\left(\sqrt[{\mathrm{3}}]{\mathrm{2}}−\mathrm{1}\right)\left(\sqrt[{\mathrm{3}}]{\mathrm{2}}+\mathrm{1}\right)^{\mathrm{3}} } \\ $$$$=\sqrt[{\mathrm{3}}]{\frac{\mathrm{1}}{\mathrm{3}}\left(\sqrt[{\mathrm{3}}]{\mathrm{4}}−\mathrm{1}\right)\left(\sqrt[{\mathrm{3}}]{\mathrm{2}}+\mathrm{1}\right)^{\mathrm{2}} }=\sqrt[{\mathrm{3}}]{\frac{\mathrm{1}}{\mathrm{3}}\left(\sqrt[{\mathrm{3}}]{\mathrm{4}}−\mathrm{1}\right)\left(\sqrt[{\mathrm{3}}]{\mathrm{4}}+\mathrm{2}\sqrt[{\mathrm{3}}]{\mathrm{2}}+\mathrm{1}\right)}…