Question Number 195364 by Rodier97 last updated on 01/Aug/23 $$ \\ $$$$\mathrm{show}\:\mathrm{that}\:\mathrm{for}\:\mathrm{any}\:\mathrm{natural}\:\mathrm{number}\:{n},\: \\ $$$$\mathrm{the}\:\mathrm{natural}\:\mathrm{number}\:\left(\mathrm{3}−\sqrt{\mathrm{5}}\right)^{{n}} +\left(\mathrm{3}+\sqrt{\mathrm{5}}\right)^{{n}} \:\mathrm{is}\:\mathrm{divisible} \\ $$$$\mathrm{by}\:\mathrm{2}^{{n}} . \\ $$ Answered by Frix last…
Question Number 195290 by Matica last updated on 29/Jul/23 $$\:{A}\:{professor}\:{said}\:\:\mathrm{0}\mid\mathrm{0}\:{because}\:\mathrm{0}=\:\mathrm{0}×{a}+\mathrm{0}\:\:\:,\:{a}\in\:\mathbb{N}.\:{Can}\:{you}\:{prove}? \\ $$ Answered by Frix last updated on 29/Jul/23 $${a}\mid{b}\:\Leftrightarrow\:\frac{{b}}{{a}}\in\mathbb{Z}\:\mathrm{but}\:\frac{\mathrm{0}}{\mathrm{0}}\:\mathrm{is}\:\mathrm{not}\:\mathrm{defined}\:\Rightarrow\:\mathrm{0}\mid\mathrm{0}\:\mathrm{is}\:\mathrm{meaningless} \\ $$ Terms of Service…
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Question Number 195232 by CrispyXYZ last updated on 28/Jul/23 $${a},\:{b},\:{c}>\mathrm{0},\:\:{a}+{b}+{c}\geqslant\mathrm{1}.\:\mathrm{Find} \\ $$$$\mathrm{max}\:\underset{\mathrm{cyc}} {\sum}\:\frac{{a}−{bc}}{{a}+{bc}}. \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 195206 by otchereabdullai@gmail.com last updated on 27/Jul/23 Answered by MM42 last updated on 27/Jul/23 $${p}\left({a}\right)={p}\left({b}\right)={p}\left({c}\right)=\frac{\mathrm{3}}{\mathrm{4}}\:;\:{probability}\:{of}\:{winning}\:{each}\:{race} \\ $$$$\left({i}\right)\:{p}\left({a}'\cap{b}\cap{c}'\right)=\frac{\mathrm{1}}{\mathrm{4}}×\frac{\mathrm{3}}{\mathrm{4}}×\frac{\mathrm{1}}{\mathrm{4}}=\frac{\mathrm{3}}{\mathrm{64}} \\ $$$$\left({ii}\right)\:{p}\left({a}\cap{b}\cap{c}\right)=\frac{\mathrm{27}}{\mathrm{64}} \\ $$$$\left({iii}\right)\:{p}\left(\Sigma{abc}'\right)=\mathrm{3}×\frac{\mathrm{1}}{\mathrm{4}}×\frac{\mathrm{9}}{\mathrm{16}}=\frac{\mathrm{27}}{\mathrm{64}} \\ $$…
Question Number 195223 by Rodier97 last updated on 27/Jul/23 $$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{In}=\int_{\mathrm{1}} ^{\mathrm{e}} \frac{\left({lnx}\right)^{{n}} }{{x}^{\mathrm{2}} }\:{dx} \\ $$$$\:\:\:\:\:\mathrm{using}\:\mathrm{an}\:\mathrm{enclosing}\:{lnx}\:\mathrm{on}\:\mathrm{interval}\:\left[\mathrm{1};\mathrm{e}\right]\:\mathrm{show}\:\mathrm{that}\:\forall{n}\:\in\:\mathbb{N}^{\ast} ,\:\mathrm{0}\:\leq\mathrm{In}\leq\:\mathrm{1} \\ $$$$ \\ $$$$ \\ $$$$…
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Question Number 195135 by 073 last updated on 25/Jul/23 Answered by som(math1967) last updated on 25/Jul/23 $$\:{x}^{\mathrm{3}} +\frac{\mathrm{1}}{{x}^{\mathrm{3}} }=\left({x}+\frac{\mathrm{1}}{{x}}\right)^{\mathrm{3}} −\mathrm{3}{x}.\frac{\mathrm{1}}{{x}}\left({x}+\frac{\mathrm{1}}{{x}}\right) \\ $$$$=\mathrm{3}\sqrt{\mathrm{3}}−\mathrm{3}\sqrt{\mathrm{3}}=\mathrm{0} \\ $$$$\:\frac{{x}^{\mathrm{6}} +\mathrm{1}}{{x}^{\mathrm{3}}…
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Question Number 195118 by York12 last updated on 25/Jul/23 $${a},{b},{c}>\mathrm{0}\:\&\:\frac{\mathrm{1}}{{a}}+\frac{\mathrm{1}}{{b}}+\frac{\mathrm{1}}{{c}}=\mathrm{3} \\ $$$${prove}\:{that} \\ $$$$\frac{{a}}{{b}^{\mathrm{2}} +{c}^{\mathrm{2}} }+\frac{{b}}{{a}^{\mathrm{2}} +{c}^{\mathrm{2}} }+\frac{{c}}{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }\geqslant\frac{\mathrm{3}}{\mathrm{2}}\left(\frac{{a}+{b}+{c}}{{ab}+{bc}+{ac}}\right)^{\mathrm{2}} \\ $$ Commented by York12…