Question Number 201814 by Euclid last updated on 13/Dec/23 Answered by sajitha last updated on 13/Dec/23 $$\mathrm{4}^{{x}^{\mathrm{3}} } =\mathrm{4} \\ $$$${x}^{\mathrm{3}} =\mathrm{1} \\ $$$${x}^{\mathrm{3}} −\mathrm{1}=\mathrm{0}…
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Question Number 201790 by ajfour last updated on 12/Dec/23 Commented by ajfour last updated on 12/Dec/23 $${Can}\:{we}\:{find}\:{sides}\:{of}\:\Box{PQRS}\:\:{in} \\ $$$${terms}\:{of}\:{a},{b},{c}\:? \\ $$ Answered by a.lgnaoui last…
Question Number 201808 by ahmetgg last updated on 12/Dec/23 Commented by ahmetgg last updated on 12/Dec/23 Syhthetic solution please. Answered by mr W last updated on 12/Dec/23…
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}}…