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…
Question Number 173152 by AgniMath last updated on 07/Jul/22 $$\mathrm{If}\:{x}\:=\:\mathrm{2}\:+\:\mathrm{2}^{\frac{\mathrm{2}}{\mathrm{3}}} \:+\:\mathrm{2}^{\frac{\mathrm{1}}{\mathrm{3}}} \:\mathrm{then}\:\mathrm{prove}\:\mathrm{that}\: \\ $$$${x}^{\mathrm{3}} \:−\:\mathrm{6}{x}^{\mathrm{2}} \:+\:\mathrm{6}{x}\:−\:\mathrm{2}\:=\:\mathrm{0}. \\ $$ Answered by Frix last updated on 07/Jul/22…
Question Number 173121 by dragan91 last updated on 06/Jul/22 Commented by Rasheed.Sindhi last updated on 07/Jul/22 $${Someone}\:{plays}\:{with}\:{the}\:{forum} \\ $$$${like}\:{a}\:{child}!\:{S}/{he}\:{colours}\:{red}\:{to} \\ $$$${some}\:{purposeful}\:{posts}!\:{Perhaps} \\ $$$${he}\:{is}\:{actually}\:{child}! \\ $$…
Question Number 173123 by dragan91 last updated on 06/Jul/22 Commented by MJS_new last updated on 06/Jul/22 $$\mathrm{both}\:\mathrm{are}\:\mathrm{linear}\:\mathrm{for}\:{y} \\ $$$$\left(\mathrm{1}\right)\:\mathrm{solve}\:\mathrm{to}\:\begin{cases}{{y}=\mathrm{term}\:\mathrm{A}}\\{{y}=\mathrm{term}\:\mathrm{B}}\end{cases} \\ $$$$\left(\mathrm{2}\right)\:\mathrm{term}\:\mathrm{A}\:=\mathrm{term}\:\mathrm{B}\:\left(\mathrm{which}\:\mathrm{is}\:\mathrm{a}\:\mathrm{polynome}\right. \\ $$$$\:\:\:\:\:\mathrm{of}\:\mathrm{4}^{\mathrm{th}} \:\mathrm{degree} \\…
Question Number 173115 by dragan91 last updated on 06/Jul/22 $$\mathrm{a}+\mathrm{b}=\mathrm{b}+\mathrm{c}=\mathrm{a}+\mathrm{c}=\mathrm{abc} \\ $$$$\left(\mathrm{a},\mathrm{b},\mathrm{c}\right)=? \\ $$ Commented by kaivan.ahmadi last updated on 06/Jul/22 $${a}+{b}={b}+{c}\Rightarrow{a}={c} \\ $$$${a}+{b}={a}+{c}\Rightarrow{b}={c} \\…