Question Number 87492 by unknown last updated on 04/Apr/20 $$\frac{\mathrm{2}+\mathrm{3}^{\mathrm{2}} }{\mathrm{1}!+\mathrm{2}!+\mathrm{3}!+\mathrm{4}!}+\frac{\mathrm{3}+\mathrm{4}^{\mathrm{2}} }{\mathrm{2}!+\mathrm{3}!+\mathrm{4}!+\mathrm{5}!}+…+\frac{\mathrm{2013}+\mathrm{2014}^{\mathrm{2}} }{\mathrm{2012}!+\mathrm{2013}!+\mathrm{2014}!+\mathrm{2015}!} \\ $$ Answered by mind is power last updated on 04/Apr/20 $$\underset{{k}\geqslant\mathrm{2}}…
Question Number 153012 by bobhans last updated on 04/Sep/21 $$\:{Find}\:{all}\:{ordered}\:{pairs}\:{of}\:{real}\: \\ $$$$\:{numbers}\:\left({x},{y}\right)\:{for}\:{which} \\ $$$$\:\:\begin{cases}{\left(\mathrm{1}+{x}^{\mathrm{4}} \right)\left(\mathrm{1}+{x}^{\mathrm{2}} \right)\left(\mathrm{1}+{x}\right)=\mathrm{1}+{y}^{\mathrm{7}} }\\{\left(\mathrm{1}+{y}^{\mathrm{4}} \right)\left(\mathrm{1}+{y}^{\mathrm{2}} \right)\left(\mathrm{1}+{y}\right)=\mathrm{1}+{x}^{\mathrm{7}} }\end{cases} \\ $$ Commented by mr…
Question Number 153019 by mathdanisur last updated on 04/Sep/21 $$\mathrm{Determine}\:\mathrm{all}\:\mathrm{functions}\:\:\mathrm{f}:\mathbb{R}\rightarrow\left(\mathrm{1};+\infty\right) \\ $$$$\mathrm{continuous}\:\mathrm{such}\:\mathrm{that} \\ $$$$\mathrm{f}\left(\mathrm{4x}\right)\:\centerdot\:\mathrm{f}\left(\mathrm{3x}\right)\:=\:\mathrm{2}^{\boldsymbol{\mathrm{x}}} \:\:;\:\:\forall\mathrm{x}\in\mathbb{R} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 21940 by Tinkutara last updated on 07/Oct/17 $$\mathrm{A}\:\mathrm{polynomial}\:\mathrm{function}\:{f}\left({x}\right)\:\mathrm{satisfies} \\ $$$${f}\left({x}\right){f}\left(\frac{\mathrm{1}}{{x}}\right)\:=\:\mathrm{2}{f}\left({x}\right)\:+\:\mathrm{2}{f}\left(\frac{\mathrm{1}}{{x}}\right);\:{x}\:\neq\:\mathrm{0}\:\mathrm{and} \\ $$$${f}\left(\mathrm{2}\right)\:=\:\mathrm{18},\:\mathrm{then}\:{f}\left(\mathrm{3}\right)\:\mathrm{is}\:\mathrm{equal}\:\mathrm{to} \\ $$ Commented by ajfour last updated on 07/Nov/18 $${No}\:{one}\:{solved}\:{this},\:{if}\:{correct} \\…
Question Number 152980 by mr W last updated on 03/Sep/21 Commented by mathdanisur last updated on 03/Sep/21 $$\mathrm{a}^{\mathrm{7}} +\mathrm{b}^{\mathrm{7}} +\mathrm{c}^{\mathrm{7}} \:=\:\mathrm{7abc}\left(\mathrm{ab}+\mathrm{bc}+\mathrm{ca}\right)^{\mathrm{2}} \\ $$$$\mathrm{a}^{\mathrm{4}} +\mathrm{b}^{\mathrm{4}} +\mathrm{c}^{\mathrm{4}}…
Question Number 152960 by nitchyy last updated on 03/Sep/21 $$\:\: \\ $$$$\:\:\:\:\mathrm{Find}\:\mathrm{the}\:\mathrm{values}\:\mathrm{for}\:\mathrm{b}\:\mathrm{and}\:\mathrm{c}\:\mathrm{given} \\ $$$$\:\:\:\:\mathrm{that}\:\mathrm{the}\:\mathrm{quadratic}\:\mathrm{expression} \\ $$$$\:\:\:\:{x}^{\mathrm{2}} +\mathrm{b}{x}+\mathrm{c}<\mathrm{0}\: \\ $$$$\:\:\:\:\left\{{x}:−\mathrm{1}<{x}>\mathrm{3}\right\} \\ $$$$\: \\ $$ Answered by…
Question Number 21877 by FilupS last updated on 06/Oct/17 $${x}=^{\mathrm{3}} \sqrt{\mathrm{7}+\mathrm{5}\sqrt{\mathrm{2}}}+^{\mathrm{3}} \sqrt{\mathrm{7}−\mathrm{5}\sqrt{\mathrm{2}}} \\ $$$$\: \\ $$$$\mathrm{1}.\:\mathrm{According}\:\mathrm{to}\:\mathrm{a}\:\mathrm{video},\:{x}=\mathrm{2} \\ $$$$\: \\ $$$$\mathrm{2}.\:\mathrm{According}\:\mathrm{to}\:\mathrm{WolframAlpha}, \\ $$$$\:\:\:\:\:\:\:\:\:\:\:{x}\approx\mathrm{0}.\mathrm{2071}+\mathrm{0}.\mathrm{3587}{i}\:\:\mathrm{for}\:“\mathrm{principal}\:\mathrm{root}'' \\ $$$$\mathrm{and}\:\:\:{x}=\mathrm{2}\sqrt{\mathrm{2}}\:\:\mathrm{for}\:“\mathrm{real}-\mathrm{valued}\:\mathrm{root}'' \\…
Question Number 152950 by john_santu last updated on 03/Sep/21 $$\left(\frac{\mathrm{1}}{\mathrm{2}}\right)+\left(\frac{\mathrm{1}}{\mathrm{3}}+\frac{\mathrm{2}}{\mathrm{3}}\right)+\left(\frac{\mathrm{1}}{\mathrm{4}}+\frac{\mathrm{2}}{\mathrm{4}}+\frac{\mathrm{3}}{\mathrm{4}}\right)+…+\left(\frac{\mathrm{1}}{\mathrm{8}}+\frac{\mathrm{2}}{\mathrm{8}}+…+\frac{\mathrm{7}}{\mathrm{8}}\right)=? \\ $$ Commented by MJS_new last updated on 03/Sep/21 $$\mathrm{14} \\ $$ Terms of Service…
Question Number 152946 by john_santu last updated on 03/Sep/21 $$\:\:\:\frac{\mathrm{4038}}{\mathrm{1}+\frac{\mathrm{1}}{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{6}}+\frac{\mathrm{1}}{\mathrm{10}}+\frac{\mathrm{1}}{\mathrm{15}}+…+\frac{\mathrm{1}}{\mathrm{1}+\mathrm{2}+\mathrm{3}+\mathrm{4}+…+\mathrm{2019}}}\:=? \\ $$ Commented by mathdanisur last updated on 03/Sep/21 $$\blacktriangle\:\frac{\mathrm{2}\centerdot\mathrm{2019}}{\mathrm{1}\:+\:\frac{\mathrm{1}}{\mathrm{1}\:+\:\mathrm{2}}\:+\:\frac{\mathrm{1}}{\mathrm{1}\:+\:\mathrm{2}\:+\:\mathrm{3}}\:+\:…\:+\:\frac{\mathrm{1}}{\mathrm{1}\:+\:\mathrm{2}\:+\:\mathrm{3}\:+\:…\:+\:\mathrm{2019}}} \\ $$$$\boldsymbol{\mathrm{A}}\:=\:\mathrm{1}\:+\:\frac{\mathrm{1}}{\mathrm{1}\:+\:\mathrm{2}}\:+\:\frac{\mathrm{1}}{\mathrm{1}\:+\:\mathrm{2}\:+\:\mathrm{3}}\:+\:…\:+\:\frac{\mathrm{1}}{\mathrm{1}\:+\:\mathrm{2}\:+\:…\:+\:\mathrm{2019}} \\ $$$$\boldsymbol{\mathrm{A}}\:=\:\mathrm{1}\:+\:\frac{\mathrm{2}}{\mathrm{2}\centerdot\mathrm{3}}\:+\:\frac{\mathrm{2}}{\mathrm{3}\centerdot\mathrm{4}}\:+\:…\:+\:\frac{\mathrm{2}}{\mathrm{2019}\centerdot\mathrm{2020}} \\…
Question Number 21874 by tawa tawa last updated on 05/Oct/17 $$\mathrm{Find}\:\mathrm{the}\:\mathrm{remainder}\:\mathrm{if}\:\:\:\mathrm{2}^{\mathrm{2006}} \:\mathrm{is}\:\mathrm{divided}\:\mathrm{by}\:\:\mathrm{17} \\ $$ Answered by mrW1 last updated on 06/Oct/17 $$\mathrm{2}^{\mathrm{2006}} =\mathrm{2}^{\mathrm{2}} ×\mathrm{2}^{\mathrm{2004}} =\mathrm{4}×\left(\mathrm{2}^{\mathrm{4}}…