Question Number 145198 by loveineq last updated on 03/Jul/21 $$\mathrm{Let}\:{a}\geqslant{b}\geqslant{c}\geqslant\mathrm{0}\:,\:{c}^{\mathrm{3}} +\left({a}+{b}\right)^{\mathrm{3}} \neq\mathrm{0}\:\mathrm{and}\:{a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} \:=\:\mathrm{3}.\:\mathrm{Prove}\:\mathrm{that} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\frac{\mathrm{1}}{\mathrm{2}}\:\leqslant\:\frac{{a}^{\mathrm{3}} +\left({b}+{c}\right)^{\mathrm{3}} +{b}^{\mathrm{3}} +\left({c}+{a}\right)^{\mathrm{3}} }{{c}^{\mathrm{3}} +\left({a}+{b}\right)^{\mathrm{3}} }\:\leqslant\:\mathrm{2} \\ $$$$\mathrm{Determine}\:\mathrm{when}\:\mathrm{equality}\:\mathrm{holds}.…
Question Number 145191 by mathdanisur last updated on 03/Jul/21 $${if}\:\:{q}\geqslant\mathrm{1}\:\:{and}\:\:{x}>−\mathrm{1}\:\:{then}: \\ $$$$\left(\mathrm{1}+{x}\right)^{\boldsymbol{{q}}} \:\geqslant\:\left(\mathrm{1}+{x}\right)^{\boldsymbol{{q}}−\mathrm{1}} \:+\:{x}\:\geqslant\:\mathrm{1}+{qx} \\ $$ Answered by Olaf_Thorendsen last updated on 03/Jul/21 $$\left(\mathrm{1}+{x}\right)^{{q}} \:=\:\left(\mathrm{1}+{x}\right)^{{q}−\mathrm{1}}…
Question Number 14119 by RasheedSindhi last updated on 28/May/17 $$\mathrm{For}\:\mathrm{what}\:\mathrm{values}\:\mathrm{of}\:\mathrm{n}\in\mathbb{N}, \\ $$$$\omega^{\mathrm{1}/\mathrm{n}} \:\mathrm{can}\:\mathrm{be}\:\mathrm{expressed}\:\mathrm{as}\:\omega^{\mathrm{m}} \\ $$$$\mathrm{where}\:\mathrm{m}\in\mathbb{Z}? \\ $$ Commented by prakash jain last updated on 28/May/17…
Question Number 14113 by FilupS last updated on 28/May/17 $$\mathrm{Is}\:\mathrm{this}\:\mathrm{incorrect}: \\ $$$${S}=\mathrm{2}×\mathrm{2}×… \\ $$$${S}=\mathrm{2}\left(\mathrm{2}×\mathrm{2}×…\right) \\ $$$${S}=\mathrm{2}{S} \\ $$$${S}=\mathrm{0} \\ $$$$\: \\ $$$$\mathrm{Please}\:\mathrm{explain}\:\mathrm{why} \\ $$ Commented…
Question Number 145187 by mathdanisur last updated on 03/Jul/21 $$\left(\sqrt[{\mathrm{3}}]{\mathrm{3}}\:+\:\sqrt[{\mathrm{5}}]{\mathrm{5}}\right)^{\mathrm{120}} \\ $$$${How}\:{many}\:{single}\:{units}\:{are}\:{there}\:{at} \\ $$$${the}\:{opening}\:{of}\:{the}\:{binomial} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 145186 by mathdanisur last updated on 03/Jul/21 $${f}:\mathbb{R}\rightarrow\mathbb{R} \\ $$$${f}\left({x}-\mathrm{1}\right)+{f}\left({x}+\mathrm{1}\right)=\sqrt{\mathrm{3}}\centerdot{f}\left({x}\right)\:;\:\forall{x}\in\mathbb{R} \\ $$$${find}\:\:{f}\left({x}-\mathrm{1}\right)+{f}\left({x}+\mathrm{5}\right)=? \\ $$ Commented by mathdanisur last updated on 03/Jul/21 $${a}\:{lot}\:{cool}\:{Ser}\:{thank}\:{you} \\…
Question Number 14109 by FilupS last updated on 28/May/17 $$\left({a}^{{p}} −{a}\right)\:\mathrm{mod}\:{p}\:=\:\mathrm{0} \\ $$$$\mathrm{when}\:\mathrm{is}\:\mathrm{this}\:\mathrm{true}? \\ $$$${a},{p}\in\mathbb{Z} \\ $$ Commented by ajfour last updated on 28/May/17 $${a}={p}\:\:\:\:\:{or}\:{just}…
Question Number 145174 by mathdanisur last updated on 02/Jul/21 Answered by mr W last updated on 02/Jul/21 Commented by mr W last updated on 02/Jul/21…
Question Number 79635 by TawaTawa last updated on 26/Jan/20 $$\mathrm{Sum}:\:\:\frac{\mathrm{1}}{\mathrm{1}}\:+\:\frac{\mathrm{1}}{\mathrm{1}\:+\:\mathrm{2}}\:+\:\frac{\mathrm{1}}{\mathrm{1}\:+\:\mathrm{2}\:+\:\mathrm{3}}\:+\:…\:\:+\:\:\frac{\mathrm{1}}{\mathrm{1}\:+\:\mathrm{2}\:+\:\mathrm{3}\:+\:…\:+\:\mathrm{8016}} \\ $$ Commented by mr W last updated on 26/Jan/20 $$=\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\frac{\mathrm{1}}{\mathrm{1}+\mathrm{2}+\mathrm{3}+…+{k}} \\ $$$$=\mathrm{2}\underset{{k}=\mathrm{1}}…
Question Number 145154 by mathdanisur last updated on 02/Jul/21 Terms of Service Privacy Policy Contact: info@tinkutara.com