Question Number 220160 by Nicholas666 last updated on 06/May/25 $$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{for}\:\mathrm{all}\:{x}\:,\:{y}\:\left[\mathrm{0}\:,\:\mathrm{1}\right]\:;\:\mathrm{prove}\:\mathrm{that}; \\ $$$$\:\:\left[\:\frac{\sqrt[{\mathrm{3}\:\:}]{\boldsymbol{{x}}^{\mathrm{3}} \:+\:\boldsymbol{{y}}^{\mathrm{3}} \:+\:\boldsymbol{\zeta}\left(\mathrm{3}\right)}}{\mathrm{1}\:+\:\boldsymbol{{e}}^{−\boldsymbol{{x}}^{\mathrm{2}} \boldsymbol{{y}}^{\mathrm{2}} } \:}\:+\:\frac{\sqrt[{\mathrm{4}\:\:}]{\boldsymbol{{x}}^{\mathrm{4}} +\:\boldsymbol{\Gamma}\left(\boldsymbol{{y}}+\mathrm{1}\right)}}{\left(\mathrm{1}\:+\:\boldsymbol{{y}}^{\mathrm{2}} \right)^{\mathrm{1}/\mathrm{3}} }\:+\:\frac{\boldsymbol{\mathrm{ln}}\left(\mathrm{1}\:+\:\boldsymbol{{x}}^{\mathrm{5}} \:+\:\boldsymbol{{y}}^{\mathrm{5}} \right)}{\:\sqrt{\mathrm{1}\:+\:\boldsymbol{{x}}^{\mathrm{2}} \:+\:\boldsymbol{{y}}^{\mathrm{2}}…
Question Number 220096 by Nicholas666 last updated on 05/May/25 $$ \\ $$$$\:\:\:\mathrm{let}\:{a},\:{b},\:{c},\:{d},\:{e}\:\mathrm{is}\:\mathrm{a}\:\mathrm{positive}\:\mathrm{real}\:\mathrm{numbers}\:\mathrm{and} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{K}\:=\:{a}\:+\:{b}\:+\:{c}\:+\:{d}\:+\:{e}\:+\mathrm{1}\:. \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{prove}\:\mathrm{that}; \\ $$$$\:\:\:\:\underset{\boldsymbol{{cyc}}} {\sum}\:\frac{\mathrm{1}}{\boldsymbol{{k}}−\boldsymbol{{a}}}\:<\:\frac{\mathrm{1}}{\mathrm{4}}\:\left(\frac{\sqrt[{\mathrm{4}\:\:\:\:}]{\boldsymbol{{e}}^{\mathrm{3}} \boldsymbol{{d}}^{\mathrm{3}} \boldsymbol{{c}}}}{\boldsymbol{{c}}^{\mathrm{3}/\mathrm{4}} \boldsymbol{{d}}^{\mathrm{1}/\mathrm{2}} \boldsymbol{{e}}^{\mathrm{1}/\mathrm{4}} \sqrt{\boldsymbol{{a}}}}\:+\:\frac{\sqrt[{\mathrm{4}\:\:\:}]{\boldsymbol{{d}}^{\:\mathrm{3}} \boldsymbol{{c}}^{\mathrm{2}}…
Question Number 219874 by mr W last updated on 03/May/25 $${find}\:\sqrt{\mathrm{2}^{\mathrm{6}^{\mathrm{2}^{\mathrm{1}^{\mathrm{4}^{\mathrm{4}} } } } } }=? \\ $$ Answered by fantastic last updated on 03/May/25…
Question Number 219933 by BaliramKumar last updated on 03/May/25 $$\mathrm{A}^{\mathrm{1}} \:+\:\mathrm{B}^{\mathrm{2}} \:+\:\mathrm{C}^{\mathrm{3}} \:+\:\mathrm{D}^{\mathrm{4}} \:=\:\overline {\mathrm{ABCD}} \\ $$$${find}\:\:{ABCD} \\ $$ Answered by MrGaster last updated on…
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Question Number 218889 by Spillover last updated on 17/Apr/25 Answered by mehdee7396 last updated on 17/Apr/25 $${tanx}+{cotx}=−\mathrm{1} \\ $$$${tan}^{\mathrm{3}} +{cot}^{\mathrm{3}} {x}=\left({tanx}+{cotx}\right)^{\mathrm{3}} −\mathrm{3}\left({tanx}×{cotx}\right)\left({tanx}+{cotx}\right) \\ $$$$=−\mathrm{1}+\mathrm{3}=\mathrm{2}=\mathrm{2}{m}^{\mathrm{2}} \Rightarrow{m}=\pm\mathrm{1}…
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