Question Number 194475 by BaliramKumar last updated on 08/Jul/23 $$\mathrm{How}\:\mathrm{many}\:\mathrm{sets}\:\mathrm{of}\:\mathrm{two}\:\mathrm{factors}\:\mathrm{of}\:\mathrm{720}\:\mathrm{are}\: \\ $$$$\mathrm{coprime}\:\mathrm{to}\:\mathrm{each}\:\mathrm{other}? \\ $$$$\left(\mathrm{A}\right)\:\mathrm{63}\:\:\:\:\left(\mathrm{B}\right)\:\mathrm{64}\:\:\:\:\:\left(\mathrm{C}\right)\:\mathrm{65}\:\:\:\:\:\:\:\:\:\left(\mathrm{D}\right)\:\mathrm{67} \\ $$$$ \\ $$ Commented by MM42 last updated on 08/Jul/23…
Question Number 194448 by York12 last updated on 07/Jul/23 $${If}\:{a}\:,\:{b}\:,\:{c}\:>\mathrm{0}\:,\:{such}\:{that}\:{a}+{b}+{c}=\mathrm{3} \\ $$$${prove}\:{that} \\ $$$$\frac{\mathrm{1}}{\mathrm{1}+{ab}}+\frac{\mathrm{1}}{\mathrm{1}+{ac}}+\frac{\mathrm{1}}{\mathrm{1}+{bc}}\geqslant\frac{\mathrm{9}}{\mathrm{2}\left(\sqrt{{a}}+\sqrt{{b}}+\sqrt{{c}}\right)} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 194421 by York12 last updated on 06/Jul/23 $$ \\ $$$${If}\:{a}\:,\:{b}\:,\:{c}\:>\mathrm{0}\:,\:{such}\:{that}\:{a}+{b}+{c}=\mathrm{3} \\ $$$${prove}\:{that} \\ $$$$\frac{\mathrm{1}}{\mathrm{1}+{ab}}+\frac{\mathrm{1}}{\mathrm{1}+{ac}}+\frac{\mathrm{1}}{\mathrm{1}+{bc}}\geqslant\frac{\mathrm{9}}{\mathrm{2}\left(\sqrt{{a}}+\sqrt{{b}}+\sqrt{{c}}\right)} \\ $$ Commented by York12 last updated on 07/Jul/23…
Question Number 194343 by BaliramKumar last updated on 04/Jul/23 $$\lfloor\mathrm{9}.\overset{−} {\mathrm{9}}\rfloor\:=\:? \\ $$ Answered by MM42 last updated on 04/Jul/23 $$\mathrm{9} \\ $$ Answered by…
Question Number 194312 by Mingma last updated on 03/Jul/23 Terms of Service Privacy Policy Contact: info@tinkutara.com
Let-a-b-c-be-real-positive-numbers-amp-abc-1-prove-that-ab-a-5-b-5-ab-bc-b-5-c-5-bc-ac-a-5-c-5-ac-1-
Question Number 194297 by York12 last updated on 02/Jul/23 $${Let}\:{a}\:,\:{b}\:,\:{c}\:{be}\:\:{real}\:{positive}\:{numbers}\:\&\: \\ $$$${abc}=\mathrm{1}\: \\ $$$${prove}\:{that} \\ $$$$\frac{{ab}}{{a}^{\mathrm{5}} +{b}^{\mathrm{5}} +{ab}}+\frac{{bc}}{{b}^{\mathrm{5}} +{c}^{\mathrm{5}} +{bc}}+\frac{{ac}}{{a}^{\mathrm{5}} +{c}^{\mathrm{5}} +{ac}}\leqslant\mathrm{1} \\ $$ Answered…
Question Number 194158 by Frix last updated on 28/Jun/23 $$\mathrm{Find}\:\mathrm{all}\:\mathrm{possible}\:\mathrm{solutions}: \\ $$$$\frac{\mathrm{1}}{{s}}+\frac{\mathrm{1}}{{t}}+\frac{\mathrm{1}}{{u}}+\frac{\mathrm{1}}{{v}}=\mathrm{1} \\ $$$$\mathrm{With}\:{s},\:{t},\:{u},\:{v}\:\in\mathbb{N}\:\mathrm{and}\:{s}<{t}<{u}<{v} \\ $$ Answered by AST last updated on 29/Jun/23 $${s}<{t}\Rightarrow\frac{\mathrm{1}}{{s}}>\frac{\mathrm{1}}{{t}}\Rightarrow\frac{\mathrm{4}}{{s}}>\frac{\mathrm{1}}{{s}}+\frac{\mathrm{1}}{{t}}+\frac{\mathrm{1}}{{u}}+\frac{\mathrm{1}}{{v}}=\mathrm{1}\Rightarrow{s}<\mathrm{4} \\…
Question Number 194144 by mr W last updated on 28/Jun/23 $${fill}\:{with}\:{different}\:{natural}\:{numbers}: \\ $$$$\:\:\frac{\mathrm{1}}{\mathrm{19}}=\frac{\mathrm{1}}{\left(\:\:\right)}+\frac{\mathrm{1}}{\left(\:\:\right)}+\frac{\mathrm{1}}{\left(\:\:\right)}+\frac{\mathrm{1}}{\left(\:\:\right)} \\ $$ Answered by York12 last updated on 28/Jun/23 $${egyptian}\:{fractions} \\ $$$$\frac{\mathrm{1}}{{k}}=\frac{\mathrm{1}}{{k}+\mathrm{1}}+\frac{\mathrm{1}}{{k}\left({k}+\mathrm{1}\right)}…
Question Number 194015 by Rupesh123 last updated on 26/Jun/23 Answered by witcher3 last updated on 26/Jun/23 $$\mathrm{let}\:\mathrm{f}\left(\mathrm{x}\right)=\underset{\mathrm{k}\geqslant\mathrm{1}} {\sum}\mathrm{f}_{\mathrm{k}} \mathrm{x}^{\mathrm{k}} \\ $$$${f}_{{k}+\mathrm{1}} ={f}_{{k}} +{f}_{{k}−\mathrm{1}} ;\mathrm{f}_{\mathrm{1}} =\mathrm{f}_{\mathrm{2}}…
Question Number 194021 by mr W last updated on 26/Jun/23 $${find}\:\sqrt[{\mathrm{3}}]{\mathrm{1}+\sqrt[{\mathrm{3}}]{\mathrm{2}}+\sqrt[{\mathrm{3}}]{\mathrm{4}}}=? \\ $$ Commented by BaliramKumar last updated on 26/Jun/23 $$\left(\mathrm{2}^{\frac{\mathrm{1}}{\mathrm{3}}} −\mathrm{1}\right)^{−\frac{\mathrm{1}}{\mathrm{3}}} \: \\ $$…