Question Number 145137 by loveineq last updated on 02/Jul/21 $$\mathrm{Let}\:{a}\geqslant{b}\geqslant{c}\geqslant\mathrm{0}\:\mathrm{and}\:{a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} \:=\:\mathrm{3}.\:\mathrm{Prove}\:\mathrm{that} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{a}^{\mathrm{3}} +\left({b}+{c}\right)^{\mathrm{3}} \:\leqslant\:\mathrm{9} \\ $$ Commented by justtry last updated on…
Question Number 14046 by RasheedSindhi last updated on 27/May/17 $$\mathrm{Calculate}:\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\sqrt{\omega} \\ $$ Answered by prakash jain last updated on 27/May/17 $${w}={e}^{\mathrm{2}\pi{i}/\mathrm{3}} \\ $$$${w}^{\mathrm{1}/\mathrm{2}}…
Question Number 145113 by imjagoll last updated on 02/Jul/21 Answered by liberty last updated on 02/Jul/21 $$\Rightarrow\mathrm{5}\left({x}^{\mathrm{4}} +{x}^{\mathrm{2}} +\mathrm{1}−\mathrm{2}{x}^{\mathrm{3}} +\mathrm{2}{x}^{\mathrm{2}} −\mathrm{2}{x}\right)=\left(\mathrm{9}{x}^{\mathrm{2}} +\mathrm{9}\right)\left({x}^{\mathrm{2}} −\mathrm{2}{x}+\mathrm{1}\right) \\ $$$$\Rightarrow\mathrm{5}{x}^{\mathrm{4}}…
Question Number 145114 by imjagoll last updated on 02/Jul/21 Commented by MJS_new last updated on 02/Jul/21 $$\mathrm{for}\:{x}\in\mathbb{R}\:\mathrm{I}\:\mathrm{get}\:\mathrm{2}+\sqrt{\mathrm{5}} \\ $$ Commented by imjagoll last updated on…
Question Number 14042 by RasheedSindhi last updated on 27/May/17 $$\mathrm{Can}\:\mathrm{we}\:\mathrm{express}\:\:\omega^{\mathrm{1}/\mathrm{2}} \:\mathrm{in}\:\mathrm{terms} \\ $$$$\mathrm{of}\:\mathrm{whole}\:\mathrm{powers}\:\mathrm{of}\:\omega? \\ $$ Commented by prakash jain last updated on 27/May/17 $$\sqrt{{w}}=\sqrt{\left(\mathrm{1}\right)\omega}=\sqrt{\omega^{\mathrm{3}} \centerdot\omega}=\sqrt{\omega^{\mathrm{4}}…
Question Number 145109 by mathdanisur last updated on 02/Jul/21 $${Solve}\:{the}\:{equation}: \\ $$$${cos}\left(\mathrm{6}{x}\right)−{cos}\left(\mathrm{4}{x}\right)=\mathrm{4}{y}^{\mathrm{2}} +\mathrm{4}{y}+\mathrm{3} \\ $$ Answered by mitica last updated on 02/Jul/21 $${cos}\mathrm{6}{x}−{cos}\mathrm{4}{x}\leqslant\mathrm{1}−\left(−\mathrm{1}\right)=\mathrm{2}\Rightarrow \\ $$$$\mathrm{4}{y}^{\mathrm{2}}…
Question Number 79571 by TawaTawa last updated on 26/Jan/20 $$\mathrm{Solve}\:\mathrm{for}\:\mathrm{x}: \\ $$$$\:\sqrt[{\mathrm{3}}]{\mathrm{x}\:\:+\:\:\sqrt[{\mathrm{3}}]{\mathrm{x}\:\:+\:\:\sqrt[{\mathrm{3}}]{\mathrm{x}\:\:+\:\:…}}}\:\:\:\:\:\:\:=\:\:\:\:\:\sqrt[{\mathrm{3}}]{\mathrm{x}\:\:\sqrt[{\mathrm{3}}]{\mathrm{x}\:\:\sqrt[{\mathrm{3}}]{\mathrm{x}\:\:….}}} \\ $$ Commented by MJS last updated on 26/Jan/20 $$\mathrm{the}\:\mathrm{rhs}\:\mathrm{has}\:\mathrm{no}\:\mathrm{defined}\:\mathrm{value}\:\mathrm{for}\:{x}<\mathrm{0} \\ $$$$\mathrm{i}.\mathrm{e}.\:\mathrm{try}\:{x}=−\mathrm{1} \\…
Question Number 145093 by mathdanisur last updated on 02/Jul/21 $${if}\:\:{f}\left({ax}+\mathrm{2}{b}\right)={x}\:\:{and}\:\:{f}\left(\mathrm{2}{a}\right)=\frac{{b}}{{a}} \\ $$$${find}\:\:{f}\left(\mathrm{5}{b}\right)=? \\ $$ Answered by liberty last updated on 02/Jul/21 $$\Leftrightarrow{f}\left({x}\right)=\frac{{x}−\mathrm{2}{b}}{{a}}\:\wedge\:{f}\left(\mathrm{2}{a}\right)=\frac{{b}}{{a}} \\ $$$$\Rightarrow\frac{\mathrm{2}{a}−\mathrm{2}{b}}{{a}}\:=\:\frac{{b}}{{a}}\:;\:\mathrm{3}{b}=\mathrm{2}{a}\:\&\:{a}=\frac{\mathrm{3}{b}}{\mathrm{2}} \\…
Question Number 79560 by jagoll last updated on 26/Jan/20 $$\sqrt{\mathrm{1}+\mathrm{x}}\:\leqslant\:\sqrt[{\mathrm{4}\:}]{\mathrm{5}−\mathrm{x}} \\ $$ Commented by john santu last updated on 26/Jan/20 $$\left(\mathrm{1}+\mathrm{x}\right)^{\mathrm{2}} \leqslant\mathrm{5}−\mathrm{x}\:,\:\mathrm{x}\leqslant\mathrm{5}\:\wedge\mathrm{x}\geqslant−\mathrm{1}\Rightarrow−\mathrm{1}\leqslant\mathrm{x}\leqslant\mathrm{5} \\ $$$$\mathrm{x}^{\mathrm{2}} +\mathrm{2x}+\mathrm{1}\leqslant\mathrm{5}−\mathrm{x}…
Question Number 145082 by mathdanisur last updated on 02/Jul/21 $$\underset{{n}=\mathrm{2}} {\overset{\infty} {\sum}}\left(\frac{\mathrm{1}}{\mathrm{2}{n}^{\mathrm{2}} −\mathrm{2}}\right)=? \\ $$ Answered by phally last updated on 02/Jul/21 $$=\left(−\frac{\mathrm{1}}{\mathrm{2}}\right)\underset{\mathrm{n}=\mathrm{2}} {\overset{\infty} {\sum}}\frac{\left(\mathrm{n}−\mathrm{1}\right)−\left(\mathrm{n}+\mathrm{1}\right)}{\mathrm{2}\left(\mathrm{n}−\mathrm{1}\right)\left(\mathrm{n}+\mathrm{1}\right)}…