Question Number 201882 by hardmath last updated on 14/Dec/23 Commented by hardmath last updated on 14/Dec/23 $$ \\ $$If angleABC=45°, angleADC=60°,DC = 2BD, then find…
Question Number 201846 by cortano12 last updated on 14/Dec/23 $$\:\:\: \\ $$ Commented by Frix last updated on 14/Dec/23 $$\mathrm{Seriously}??? \\ $$$$\mathrm{3184}:\mathrm{2}=\mathrm{1592} \\ $$$$\mathrm{1592}:\mathrm{2}=\mathrm{796} \\…
Question Number 201864 by ajfour last updated on 14/Dec/23 $$\left({ct}^{\mathrm{2}} −\frac{\mathrm{1}}{{ct}^{\mathrm{2}} }+\sqrt{\mathrm{2}}\right)^{\mathrm{2}} ={c}^{\mathrm{2}} \left({t}^{\mathrm{4}} +\frac{\mathrm{1}}{{t}^{\mathrm{4}} }\right) \\ $$$${Find}\:\:{t}={f}\left({c}\right). \\ $$ Commented by Frix last updated…
Question Number 201857 by cortano12 last updated on 14/Dec/23 $$\:\:\mathrm{If}\:{a},{b},{c},{d},{e}\:\mathrm{are}\:\mathrm{thr}\:\mathrm{roots}\:\mathrm{of}\: \\ $$$$\:\:\mathrm{2x}^{\mathrm{5}} −\mathrm{3x}^{\mathrm{3}} +\mathrm{2x}−\mathrm{7}=\mathrm{0}\:,\:\mathrm{find}\:\mathrm{the}\: \\ $$$$\:\mathrm{value}\:\mathrm{of}\:\underset{\mathrm{cyc}} {\prod}\left({a}^{\mathrm{3}} −\mathrm{1}\right)\: \\ $$ Commented by Frix last updated…
Question Number 201859 by York12 last updated on 14/Dec/23 $$\mathrm{If}\:{xyz}\:\in\mathbb{R}^{+} \:,\:{xyz}=\mathrm{1}\:,\:\mathrm{prove}\:\mathrm{that}\:\mathrm{the}\:\mathrm{following}\:\mathrm{inequality}\:\mathrm{holds}: \\ $$$$\frac{{x}}{\mathrm{2}{x}^{\mathrm{5}} +{x}+\mathrm{4}}+\frac{{y}}{\mathrm{2}{y}^{\mathrm{5}} +{y}+\mathrm{4}}+\frac{{z}}{\mathrm{2}{z}^{\mathrm{5}} +{z}+\mathrm{4}}\geqslant\frac{\mathrm{3}}{\mathrm{7}}. \\ $$$$\boldsymbol{\mathrm{Solution}}\:\boldsymbol{\mathrm{please}}\:\boldsymbol{\mathrm{with}}\:\boldsymbol{\mathrm{an}}\:\boldsymbol{\mathrm{advice}}\:\boldsymbol{\mathrm{to}}\:\boldsymbol{\mathrm{get}}\:\boldsymbol{\mathrm{better}} \\ $$$$\boldsymbol{\mathrm{at}}\:\boldsymbol{\mathrm{inequalities}}\:\boldsymbol{\mathrm{and}}\:\boldsymbol{\mathrm{which}}\:\boldsymbol{\mathrm{book}}\:\boldsymbol{\mathrm{would}}\:\boldsymbol{\mathrm{you}}\:\boldsymbol{\mathrm{recommend}}. \\ $$$$\boldsymbol{\mathrm{Thanks}}\:\boldsymbol{\mathrm{in}}\:\boldsymbol{\mathrm{advance}}! \\ $$$$\: \\…
Question Number 201820 by cortano12 last updated on 13/Dec/23 $$\:\:\:\frac{\mid\mathrm{3x}+\mathrm{1}\mid−\mid\mathrm{x}+\mathrm{2}\mid}{\mathrm{3}−\mid\mathrm{2x}\mid}\:\geqslant\:\mathrm{0}\: \\ $$$$\:\:\mathrm{find}\:\mathrm{the}\:\mathrm{solution}\:\mathrm{set}. \\ $$ Answered by dimentri last updated on 13/Dec/23 $$\:\:\frac{\left(\mathrm{3x}+\mathrm{1}\right)^{\mathrm{2}} −\left(\mathrm{x}+\mathrm{2}\right)^{\mathrm{2}} }{\mathrm{3}^{\mathrm{2}} −\left(\mathrm{2x}\right)^{\mathrm{2}}…
Question Number 201834 by sebastian last updated on 13/Dec/23 $$ \\ $$$$ \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 201819 by York12 last updated on 13/Dec/23 $$\mathrm{If}\:{xyz}\:\in\mathbb{R}^{+} \:,\:{xyz}=\mathrm{1}\:,\:\mathrm{prove}\:\mathrm{that}\:\mathrm{the}\:\mathrm{following}\:\mathrm{inequality}\:\mathrm{holds}: \\ $$$$\frac{{x}}{\mathrm{2}{x}^{\mathrm{5}} +{x}+\mathrm{4}}+\frac{{y}}{\mathrm{2}{y}^{\mathrm{5}} +{y}+\mathrm{4}}+\frac{{z}}{\mathrm{2}{z}^{\mathrm{5}} +{z}+\mathrm{4}}\geqslant\frac{\mathrm{3}}{\mathrm{7}}. \\ $$$$\boldsymbol{\mathrm{Solution}}\:\boldsymbol{\mathrm{please}}\:\boldsymbol{\mathrm{with}}\:\boldsymbol{\mathrm{an}}\:\boldsymbol{\mathrm{advice}}\:\boldsymbol{\mathrm{to}}\:\boldsymbol{\mathrm{get}}\:\boldsymbol{\mathrm{better}} \\ $$$$\boldsymbol{\mathrm{at}}\:\boldsymbol{\mathrm{inequalities}}\:\boldsymbol{\mathrm{and}}\:\boldsymbol{\mathrm{which}}\:\boldsymbol{\mathrm{book}}\:\boldsymbol{\mathrm{would}}\:\boldsymbol{\mathrm{you}}\:\boldsymbol{\mathrm{recommend}}. \\ $$$$\boldsymbol{\mathrm{Thanks}}\:\boldsymbol{\mathrm{in}}\:\boldsymbol{\mathrm{advance}}! \\ $$$$\: \\…
Question Number 201804 by York12 last updated on 12/Dec/23 $$\mathrm{If}\:{xyz}\:\in\mathbb{R}^{+} \:,\:{xyz}=\mathrm{1}\:,\:\mathrm{prove}\:\mathrm{that}\:\mathrm{the}\:\mathrm{following}\:\mathrm{inequality}\:\mathrm{holds}: \\ $$$$\frac{{x}}{\mathrm{2}{x}^{\mathrm{5}} +{x}+\mathrm{4}}+\frac{{y}}{\mathrm{2}{y}^{\mathrm{5}} +{y}+\mathrm{4}}+\frac{{z}}{\mathrm{2}{z}^{\mathrm{5}} +{z}+\mathrm{4}}\geqslant\frac{\mathrm{3}}{\mathrm{7}}. \\ $$$$\boldsymbol{\mathrm{Solution}}\:\boldsymbol{\mathrm{please}}\:\boldsymbol{\mathrm{with}}\:\boldsymbol{\mathrm{an}}\:\boldsymbol{\mathrm{advice}}\:\boldsymbol{\mathrm{to}}\:\boldsymbol{\mathrm{get}}\:\boldsymbol{\mathrm{better}} \\ $$$$\boldsymbol{\mathrm{at}}\:\boldsymbol{\mathrm{inequalities}}\:\boldsymbol{\mathrm{and}}\:\boldsymbol{\mathrm{which}}\:\boldsymbol{\mathrm{book}}\:\boldsymbol{\mathrm{would}}\:\boldsymbol{\mathrm{you}}\:\boldsymbol{\mathrm{recommend}}. \\ $$$$\boldsymbol{\mathrm{Thanks}}\:\boldsymbol{\mathrm{in}}\:\boldsymbol{\mathrm{advance}}! \\ $$$$\: \\…
Question Number 201802 by Frix last updated on 12/Dec/23 $$\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}} \\…