Question Number 107674 by Study last updated on 12/Aug/20 $${f}\left({x}\right)=\sqrt{{x}}\sqrt{{x}}\:\:\:\:\:\:\:{D}_{{f}} =??? \\ $$ Answered by Her_Majesty last updated on 12/Aug/20 $${x}={re}^{{i}\theta} ;\:{r}\in\mathbb{R}^{+} \wedge\theta\in\mathbb{R} \\ $$$$\Rightarrow…
Question Number 42133 by rahul 19 last updated on 18/Aug/18 $$\mathrm{Let}\:\mathrm{a},\mathrm{b},\mathrm{c}\epsilon\:\mathrm{R}\:\mathrm{such}\:\mathrm{that}\:\mathrm{a}+\mathrm{2b}+\mathrm{c}=\mathrm{4}. \\ $$$$\mathrm{Find}\:\mathrm{maximum}\:\mathrm{value}\:\mathrm{of}\:\left(\mathrm{ab}+\mathrm{bc}+\mathrm{ca}\right). \\ $$ Answered by MrW3 last updated on 18/Aug/18 $${b}=\frac{\mathrm{4}−{a}−{c}}{\mathrm{2}}=\mathrm{2}−\frac{{a}+{c}}{\mathrm{2}} \\ $$$${F}\left({a},{c}\right)={ab}+{bc}+{ca}=\left({a}+{c}\right){b}+{ac}=\mathrm{2}\left({a}+{c}\right)−\frac{\left({a}+{c}\right)^{\mathrm{2}}…
Question Number 107673 by qwerty111 last updated on 12/Aug/20 Answered by Her_Majesty last updated on 12/Aug/20 $$=\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\left(\mathrm{2}^{\mathrm{2}{k}} +\mathrm{2}^{−\mathrm{2}{k}} +\mathrm{2}\right)=\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\mathrm{4}^{{k}} +\underset{{k}=\mathrm{1}} {\overset{{n}}…
Question Number 173196 by AgniMath last updated on 08/Jul/22 $$\mathrm{What}\:\mathrm{is}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of} \\ $$$$\frac{{y}^{\mathrm{2}} \:+\:{yz}\:+\:{z}^{\mathrm{2}} }{\left({x}\:−\:{y}\right)\left({x}\:−\:{z}\right)}+\frac{{z}^{\mathrm{2}} \:+\:{zx}\:+\:{x}^{\mathrm{2}} }{\left({y}\:−\:{z}\right)\left({z}\:−\:{y}\right)}+\frac{{x}^{\mathrm{2}} \:+\:{xy}\:+\:{y}^{\mathrm{2}} }{\left({z}\:−\:{x}\right)\left({z}−{y}\right)} \\ $$ Commented by som(math1967) last updated…
Question Number 42118 by Akashuac last updated on 18/Aug/18 $$\mathrm{x}−\mathrm{4}=−\frac{\mathrm{1}}{\mathrm{x}}\:\mathrm{then}\:.\:\mathrm{prove}\:\mathrm{x}^{\mathrm{4}} −\mathrm{194}=−\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{4}} } \\ $$ Answered by tanmay.chaudhury50@gmail.com last updated on 18/Aug/18 $${x}+\frac{\mathrm{1}}{{x}}=\mathrm{4} \\ $$$${x}^{\mathrm{2}} +\frac{\mathrm{1}}{{x}^{\mathrm{2}}…
Question Number 173184 by Shrinava last updated on 07/Jul/22 $$\mathrm{Solve}\:\mathrm{for}\:\mathrm{real}\:\mathrm{numbers}: \\ $$$$\mathrm{2}\:\int_{\mathrm{0}} ^{\:\boldsymbol{\mathrm{x}}} \:\frac{\mathrm{x}^{\mathrm{2}} \:\centerdot\:\mathrm{e}^{\boldsymbol{\mathrm{arctan}}\left(\boldsymbol{\mathrm{x}}\right)} }{\:\sqrt{\mathrm{1}\:+\:\mathrm{x}^{\mathrm{2}} }}\:\mathrm{dx}\:=\:\mathrm{1} \\ $$ Answered by aleks041103 last updated on…
Question Number 173171 by AgniMath last updated on 07/Jul/22 $$\mathrm{Find}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:{x}\:\mathrm{from}\:\mathrm{the}\:\mathrm{following} \\ $$$$\mathrm{equation}: \\ $$$$\left({x}\:−\:\mathrm{3}\right)\left({x}\:−\:\mathrm{4}\right)\:=\:\frac{\mathrm{34}}{\mathrm{33}^{\mathrm{2}} } \\ $$ Commented by som(math1967) last updated on 08/Jul/22 $${x}=\mathrm{2}\frac{\mathrm{32}}{\mathrm{33}}\:\:,\mathrm{4}\frac{\mathrm{1}}{\mathrm{33}}…
Question Number 173163 by dragan91 last updated on 07/Jul/22 Answered by mr W last updated on 07/Jul/22 $${y}=\pm{x} \\ $$$$\left({x}−\mathrm{2}\left({a}−\mathrm{1}\right)\right)^{\mathrm{2}} +\left({x}+{a}\right)^{\mathrm{2}} =\mathrm{1} \\ $$$$\mathrm{2}{x}^{\mathrm{2}} −\mathrm{2}\left({a}−\mathrm{2}\right){x}+\left({a}−\mathrm{1}\right)\left(\mathrm{5}{a}−\mathrm{3}\right)=\mathrm{0}…
Question Number 173162 by dragan91 last updated on 07/Jul/22 Answered by aleks041103 last updated on 07/Jul/22 $${x}+{y}>\mathrm{0} \\ $$$$\Rightarrow{x}^{{z}} >{y}^{{z}} \\ $$$${z}\in\mathbb{N}\Rightarrow{x}>{y} \\ $$$${let}\:{x}={y}+{t},\:{t}\in\mathbb{N} \\…
Question Number 173157 by AgniMath last updated on 07/Jul/22 $$\mathrm{If}\:\frac{\mathrm{1}}{{a}}\:+\:\frac{\mathrm{1}}{{b}}\:+\:\frac{\mathrm{1}}{{c}}\:=\:\frac{\mathrm{1}}{{a}\:+\:{b}\:+\:{c}}\:\mathrm{then}\:\mathrm{prove}\:\mathrm{that} \\ $$$$\frac{\mathrm{1}}{{a}^{\mathrm{3}} }\:+\:\frac{\mathrm{1}}{{b}^{\mathrm{3}} }\:+\:\frac{\mathrm{1}}{{c}^{\mathrm{3}} }\:=\:\frac{\mathrm{1}}{\left({a}\:+\:{b}\:+\:{c}\right)^{\mathrm{3}} }\:. \\ $$ Answered by Cesar1994 last updated on 07/Jul/22…