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Category: Arithmetic

Question-220737

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Question Number 220737 by Spillover last updated on 18/May/25 Answered by som(math1967) last updated on 18/May/25 $$\:\mathrm{0}.\overset{.} {\mathrm{6}}=\mathrm{0}.\mathrm{6}+\mathrm{0}.\mathrm{06}+\mathrm{0}.\mathrm{006}+… \\ $$$$=\frac{\mathrm{0}.\mathrm{6}}{\mathrm{1}−.\mathrm{1}}=\frac{\mathrm{6}}{\mathrm{9}}=\frac{\mathrm{2}}{\mathrm{3}} \\ $$ Commented by Spillover…

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Question Number 220159 by Nicholas666 last updated on 06/May/25 $$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{for}\:\mathrm{all}\:{x},\:{y}\:\in\:\left[\mathrm{0}\:,\:\mathrm{1}\right]\:;\:\mathrm{prove}\:\mathrm{that}; \\ $$$$\:\:\:\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}\:+\:{x}^{\mathrm{4}} }}\:+\:\frac{\mathrm{2}}{\:\sqrt{\mathrm{1}\:+\:{y}^{\mathrm{4}} }}\:+\:\frac{\mathrm{2}}{\:\sqrt{\mathrm{4}\:+\:\left({x}\:+\:{y}\right)^{\mathrm{4}} }}\:+\:\frac{\mathrm{2}\sqrt{\mathrm{2}}}{\:\sqrt{\mathrm{2}+\:{x}^{\mathrm{2}} {y}^{\mathrm{2}} \:+\:{y}^{\mathrm{3}} }}\:\:\:\:\:\:\:\: \\ $$$$\:\:\:\:\leqslant\:\frac{\mathrm{2}}{\:\sqrt{\mathrm{1}\:+\:{x}^{\mathrm{2}} {y}^{\mathrm{2}} }}\:+\:\frac{\mathrm{2}}{\:^{\mathrm{4}} \sqrt{\mathrm{1}\:+\:{x}^{\mathrm{5}}…

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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}}…

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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}}…

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