Question Number 64284 by Tawa1 last updated on 16/Jul/19 $$\mathrm{Find}\:\mathrm{all}\:\mathrm{the}\:\mathrm{integer}\:\mathrm{solution}\:\:\:\:\:\:\:\mathrm{3x}^{\mathrm{2}} \:+\:\mathrm{1}\:\:=\:\:\mathrm{4y}^{\mathrm{3}} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 64187 by Hope last updated on 15/Jul/19 Answered by Hope last updated on 15/Jul/19 $$\frac{{yz}}{{xyz}}+\frac{\mathrm{4}{zx}}{{xyz}}+\frac{\mathrm{9}{xy}}{{xyz}} \\ $$$$\frac{\mathrm{1}^{\mathrm{2}} }{{x}}+\frac{\mathrm{2}^{\mathrm{2}} }{{y}}+\frac{\mathrm{3}^{\mathrm{2}} }{{z}}\geqslant\frac{\left(\mathrm{1}+\mathrm{2}+\mathrm{3}\right)^{\mathrm{2}} }{\left({x}+{y}+{z}\right)} \\ $$$$\frac{\mathrm{1}^{\mathrm{2}}…
Question Number 64186 by Hope last updated on 15/Jul/19 Answered by Hope last updated on 15/Jul/19 $$\frac{\left(\sqrt{\mathrm{2}}\:\right)^{\mathrm{2}} }{{x}+{y}}+\frac{\left(\sqrt{\mathrm{2}}\:\right)^{\mathrm{2}} }{{y}+{z}}+\frac{\left(\sqrt{\mathrm{2}}\:\right)^{\mathrm{2}} }{{z}+{x}}\geqslant\frac{\left(\sqrt{\mathrm{2}}\:+\sqrt{\mathrm{2}}\:+\sqrt{\mathrm{2}}\:\right)^{\mathrm{2}} }{\mathrm{2}\left({x}+{y}+{z}\right)} \\ $$$${do}\geqslant\frac{\mathrm{9}×\mathrm{2}}{\mathrm{2}\left({x}+{y}+{z}\right)} \\ $$$$\frac{\mathrm{2}}{{x}+{y}}+\frac{\mathrm{2}}{{y}+{z}}+\frac{\mathrm{2}}{{z}+{x}}\geqslant\frac{\mathrm{9}}{{x}+{y}+{z}}\:{proved}…
Question Number 64185 by Hope last updated on 15/Jul/19 Commented by Hope last updated on 15/Jul/19 $${all}\:{question}\:{based}\:{on}\:{Tito}−{lemma}\:{inequality} \\ $$ Commented by Tony Lin last updated…
Question Number 64174 by gajrajgchouhan last updated on 15/Jul/19 $$\mathrm{2}^{{n}} \:/\:\mathrm{5}^{\mathrm{2}^{{n}} } \:+\:\mathrm{1}\:{infinite}\:{series}\:{sum}\:{ffom}\:\mathrm{0}\:{to}\: \\ $$$${infinity} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 129687 by ajfour last updated on 17/Jan/21 $$\:\:\:\:{x}^{\mathrm{3}} −{x}−{c}=\mathrm{0} \\ $$$$\left({solve}\:{for}\:{x}\:\:{even}\:{if}\:\:\:\mathrm{0}<{c}<\frac{\mathrm{2}}{\mathrm{3}\sqrt{\mathrm{3}}}\right) \\ $$ Answered by ajfour last updated on 18/Jan/21 $${Just}\:{now}\:{I}\:{developed}\:{a}\:{new} \\ $$$${approximate}\:{formula}:…
Question Number 129677 by Adel last updated on 17/Jan/21 Commented by som(math1967) last updated on 17/Jan/21 $$\mathrm{3}^{\frac{\mathrm{1}}{{cscx}}×\frac{\mathrm{1}}{{sinx}}×\frac{\mathrm{1}}{{tanx}}×\frac{\mathrm{1}}{{cotx}}} =\mathrm{3}^{\mathrm{1}} =\mathrm{3} \\ $$ Commented by Adel last…
Question Number 64130 by ANTARES VY last updated on 14/Jul/19 $$\sqrt{\mathrm{4}\boldsymbol{\mathrm{x}}+\frac{\mathrm{12}}{\boldsymbol{\mathrm{x}}}}=\frac{\boldsymbol{\mathrm{x}}^{\mathrm{2}} +\mathrm{7}}{\boldsymbol{\mathrm{x}}+\mathrm{1}} \\ $$$$\boldsymbol{\mathrm{x}}=? \\ $$ Commented by Prithwish sen last updated on 14/Jul/19 $$\mathrm{x}=\:\mathrm{1},\mathrm{1},\mathrm{3}\:\mathrm{and}\:\frac{−\mathrm{1}\pm\mathrm{15i}}{\mathrm{2}}…
Question Number 64126 by Cheyboy last updated on 13/Jul/19 Answered by ajfour last updated on 13/Jul/19 $$\:\:\mathrm{32}\:\:\:\:\:\:\:\:\:\:\:\mathrm{0}\:\:\:\:\:\:\:\:\:\:−\mathrm{8}\:\:\:\:\:\:\:\:\:−\mathrm{10}\:\:\:\:\:\:\:\:\:\:\:\:\:\:−\mathrm{24} \\ $$$$\:\:\:\:\:\:\:\:\:−\mathrm{32}\:\:\:\:\:−\mathrm{8}\:\:\:\:\:\:\:\:\:−\mathrm{2}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:−\mathrm{14} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{24}\:\:\:\:\:\:\:\:\:\:\mathrm{6}\:\:\:\:\:\:\:\:\:\:\:\:\:\:−\mathrm{12} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:−\mathrm{18}\:\:\:\:\:\:\:−\mathrm{18} \\ $$$${d}_{{n}}…
Question Number 129663 by liberty last updated on 17/Jan/21 $$\:\sqrt{\mathrm{14}+\sqrt{\mathrm{3}}\:+\sqrt[{\mathrm{3}}]{\mathrm{26}−\mathrm{5}\sqrt{\mathrm{3}}}}\:=?\: \\ $$ Commented by MathSh last updated on 17/Jan/21 $$\sqrt[{\mathrm{3}}]{\left(\mathrm{26}−\mathrm{15}\sqrt{\mathrm{3}}\right)^{\mathrm{3}} }=\sqrt[{\mathrm{3}}]{\left(\mathrm{2}−\sqrt{\mathrm{3}}\right)^{\mathrm{3}} }=\mathrm{2}−\sqrt{\mathrm{3}} \\ $$$$\sqrt{\mathrm{14}+\sqrt{\mathrm{3}}+\mathrm{2}−\sqrt{\mathrm{3}}}=\sqrt{\mathrm{16}}=\mathrm{4} \\…