Question Number 201804 by York12 last updated on 12/Dec/23 $$\mathrm{If}\:{xyz}\:\in\mathbb{R}^{+} \:,\:{xyz}=\mathrm{1}\:,\:\mathrm{prove}\:\mathrm{that}\:\mathrm{the}\:\mathrm{following}\:\mathrm{inequality}\:\mathrm{holds}: \\ $$$$\frac{{x}}{\mathrm{2}{x}^{\mathrm{5}} +{x}+\mathrm{4}}+\frac{{y}}{\mathrm{2}{y}^{\mathrm{5}} +{y}+\mathrm{4}}+\frac{{z}}{\mathrm{2}{z}^{\mathrm{5}} +{z}+\mathrm{4}}\geqslant\frac{\mathrm{3}}{\mathrm{7}}. \\ $$$$\boldsymbol{\mathrm{Solution}}\:\boldsymbol{\mathrm{please}}\:\boldsymbol{\mathrm{with}}\:\boldsymbol{\mathrm{an}}\:\boldsymbol{\mathrm{advice}}\:\boldsymbol{\mathrm{to}}\:\boldsymbol{\mathrm{get}}\:\boldsymbol{\mathrm{better}} \\ $$$$\boldsymbol{\mathrm{at}}\:\boldsymbol{\mathrm{inequalities}}\:\boldsymbol{\mathrm{and}}\:\boldsymbol{\mathrm{which}}\:\boldsymbol{\mathrm{book}}\:\boldsymbol{\mathrm{would}}\:\boldsymbol{\mathrm{you}}\:\boldsymbol{\mathrm{recommend}}. \\ $$$$\boldsymbol{\mathrm{Thanks}}\:\boldsymbol{\mathrm{in}}\:\boldsymbol{\mathrm{advance}}! \\ $$$$\: \\…
Question Number 201806 by sonukgindia last updated on 12/Dec/23 Commented by mr W last updated on 30/Dec/23 $${see}\:{Q}\mathrm{202543} \\ $$$${answer}\:{is}\:\mathrm{2}\left(\mathrm{4}^{\mathrm{3}} −\sqrt{\mathrm{1}}−\sqrt{\mathrm{2}}−\sqrt{\mathrm{3}}−…−\sqrt{\mathrm{15}}\right) \\ $$ Commented by…
Question Number 201802 by Frix last updated on 12/Dec/23 $$\mathrm{Solution}\:\mathrm{of}\:\mathrm{equations}\:\mathrm{like}\:\mathrm{this}: \\ $$$$\sqrt[{{n}}]{{f}\left({x}\right)}+\sqrt[{{n}}]{{g}\left({x}\right)}=\sqrt[{{n}}]{{h}\left({x}\right)} \\ $$$$\mathrm{If}\:\mathrm{the}\:\mathrm{solution}\:\mathrm{is}\:\mathrm{not}\:\mathrm{obvious}\:\mathrm{we}\:\mathrm{must}\:\mathrm{get} \\ $$$$\mathrm{rid}\:\mathrm{of}\:\mathrm{the}\:\mathrm{radicals}.\:\mathrm{In}\:\mathrm{the}\:\mathrm{following}\:\mathrm{cases} \\ $$$$\mathrm{this}\:\mathrm{is}\:\mathrm{easy}: \\ $$$$\sqrt{{a}}+\sqrt{{b}}=\sqrt{{c}} \\ $$$$\:\:\:\:\:\Rightarrow\:{a}+\mathrm{2}\sqrt{{ab}}+{b}={c} \\ $$$$\:\:\:\:\:\:\:\:\:\:\Rightarrow\:\mathrm{4}{ab}=\left({c}−{a}−{b}\right)^{\mathrm{2}} \\…
Question Number 201796 by ajfour last updated on 12/Dec/23 $${x}^{\mathrm{4}} −\mathrm{15}{x}^{\mathrm{2}} −\mathrm{30}{x}+\mathrm{104}=\mathrm{0} \\ $$$${for}\:{x}\in\mathbb{R}\:\:\:\:\:{x}=\mathrm{2},\:\mathrm{4} \\ $$$${I}\:{want}\:{to}\:{know}\:{the}\:{best}\:{way}\:{to} \\ $$$${arrive}\:{at}\:{these}\:{answers}\:\left({without}\right. \\ $$$$\left.{guessing}\right).\:{I}\:{found}\:{one}\:{new}\:{way}. \\ $$$$\:{Shall}\:{post}\:{later}. \\ $$ Commented…
Question Number 201798 by hardmath last updated on 12/Dec/23 $$\mathrm{Find}\:\mathrm{the}\:\mathrm{differential}\:\mathrm{of}\:\mathrm{the}\:\mathrm{function}: \\ $$$$\boldsymbol{\mathrm{y}}\:=\:\sqrt{\boldsymbol{\mathrm{x}}^{\mathrm{2}} \:−\:\mathrm{1}} \\ $$ Answered by dimentri last updated on 12/Dec/23 $$\:{y}=\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}\:\Rightarrow{y}'=\frac{{x}}{\:\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}}=\frac{{x}}{{y}}…
Question Number 201793 by sonukgindia last updated on 12/Dec/23 Answered by mr W last updated on 12/Dec/23 Commented by mr W last updated on 12/Dec/23…
Question Number 201724 by LimPorly last updated on 11/Dec/23 $$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{{e}^{{x}} \left({x}−\mathrm{2}\right)+{x}+\mathrm{2}}{{x}^{\mathrm{3}} }\:{solve}\:{it}\:{by}\:{not}\:{using} \\ $$$${taylor}\:{series}\:{or}\:{l}'{hopital}\:{rule}. \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 201727 by Spillover last updated on 11/Dec/23 $$\int_{\frac{\pi}{\mathrm{6}}} ^{\frac{\pi}{\mathrm{3}}} {e}^{\mathrm{sin}\:{x}^{{c}\mathrm{os}\:{x}^{\mathrm{tan}\:{x}^{\mathrm{cot}\:{x}^{\mathrm{sec}\:{x}^{\mathrm{cosec}\:{x}} } } } } } {dx} \\ $$ Answered by MathematicalUser2357 last updated…
Question Number 201722 by Calculusboy last updated on 11/Dec/23 Commented by Frix last updated on 11/Dec/23 $$\mathrm{There}\:\mathrm{also}\:\mathrm{should}\:\mathrm{be}\:\mathrm{complex}\:\mathrm{solutions} \\ $$ Answered by Sutrisno last updated on…
Question Number 201751 by esmaeil last updated on 11/Dec/23 $${hi} \\ $$$${Is}\:\:{it}\:\:{possible}\:{drawing}\:{with}\:{naming} \\ $$$${by}\:{math}\:{editor}? \\ $$$$\left({sorry}\:{i}'{m}\:{not}\:{good}\:{in}\:{english}\right) \\ $$ Commented by Rasheed.Sindhi last updated on 11/Dec/23…