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Category: Algebra

Question-80178

Question Number 80178 by Power last updated on 31/Jan/20 Commented by mr W last updated on 31/Jan/20 $${no}\:{exact}\:{value}\:{possible}! \\ $$$${x}^{\mathrm{25}} \underset{{k}=\mathrm{1}} {\overset{\mathrm{25}} {\prod}}\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{3}^{{k}} }\right)=\mathrm{1} \\…

solve-for-x-x-1-x-1-ax-2-x-a-2-1-a-R-

Question Number 80144 by behi83417@gmail.com last updated on 31/Jan/20 $$\boldsymbol{\mathrm{solve}}\:\boldsymbol{\mathrm{for}}\:\boldsymbol{\mathrm{x}}: \\ $$$$\frac{\sqrt{\boldsymbol{\mathrm{x}}}+\mathrm{1}}{\:\sqrt{\boldsymbol{\mathrm{x}}+\mathrm{1}}}+\boldsymbol{\mathrm{ax}}^{\mathrm{2}} =\boldsymbol{\mathrm{x}}\left(\boldsymbol{\mathrm{a}}^{\mathrm{2}} +\mathrm{1}\right)\:\:\:\:\:\:\left[\boldsymbol{\mathrm{a}}\in\boldsymbol{\mathrm{R}}\right] \\ $$ Commented by john santu last updated on 31/Jan/20 $${let}\:\sqrt{{x}+\mathrm{1}}\:=\:\mathrm{sec}\:{t}…

3-360-1-4-2-162-1-3-10-2-

Question Number 145683 by imjagoll last updated on 07/Jul/21 $$\:\frac{\mathrm{3}\:\sqrt[{\sqrt{\mathrm{4}}}]{\mathrm{360}}\:−\mathrm{2}\:\sqrt[{!\mathrm{3}}]{\mathrm{162}}}{\:\sqrt{\mathrm{10}}−\sqrt{\mathrm{2}}}\:=? \\ $$ Answered by puissant last updated on 07/Jul/21 $$\sqrt{\mathrm{4}}=\mathrm{2}\:\:;\:\:!\mathrm{3}=\:\:\mathrm{3}!\mid\frac{\mathrm{1}!}{\mathrm{0}!}−\frac{\mathrm{1}!}{\mathrm{1}!}−\frac{\mathrm{1}!}{\mathrm{2}!}+\frac{\mathrm{1}!}{\mathrm{3}!}\mid \\ $$$$\Rightarrow\frac{\mathrm{3}\sqrt{\mathrm{360}}−\mathrm{2}\sqrt{\mathrm{162}}}{\:\sqrt{\mathrm{10}}−\sqrt{\mathrm{2}}}=\frac{\mathrm{18}\sqrt{\mathrm{10}}−\mathrm{18}\sqrt{\mathrm{2}}}{\:\sqrt{\mathrm{10}}−\sqrt{\mathrm{2}}} \\ $$$$=\frac{\mathrm{18}\left(\sqrt{\mathrm{10}}−\sqrt{\mathrm{2}}\right)}{\left(\sqrt{\mathrm{10}}−\sqrt{\mathrm{2}}\right)}\:=\:\mathrm{18}.. \\…

a-k-1-k-3-2-k-b-k-1-k-3-k-2-k-1-7-k-

Question Number 80142 by behi83417@gmail.com last updated on 31/Jan/20 $$\mathrm{a}.\:\:\:\underset{\boldsymbol{\mathrm{k}}=\mathrm{1}} {\overset{\infty} {\sum}}\:\:\left(\frac{\boldsymbol{\mathrm{k}}^{\mathrm{3}} }{\mathrm{2}^{\boldsymbol{\mathrm{k}}} }\right)=? \\ $$$$\boldsymbol{\mathrm{b}}.\:\:\:\:\underset{\boldsymbol{\mathrm{k}}=\mathrm{1}} {\overset{\infty} {\sum}}\:\:\left(\frac{\boldsymbol{\mathrm{k}}^{\mathrm{3}} +\boldsymbol{\mathrm{k}}^{\mathrm{2}} +\boldsymbol{\mathrm{k}}+\mathrm{1}}{\mathrm{7}^{\boldsymbol{\mathrm{k}}} }\right)=? \\ $$ Answered by…