Question Number 94521 by Abdulrahman last updated on 19/May/20 $$\mathrm{if}\:\mathrm{a}^{\mathrm{10}} +\mathrm{a}^{\mathrm{5}} +\mathrm{1}=\mathrm{0} \\ $$$$\mathrm{then}\:\mathrm{a}^{\mathrm{2005}} +\frac{\mathrm{1}}{\mathrm{a}^{\mathrm{2005}} }=? \\ $$$$\mathrm{a}:\:\mathrm{a}^{\mathrm{10}} +\mathrm{a}^{\mathrm{11}} \:\:\:\:\mathrm{b}:\mathrm{a}^{\mathrm{10}} +\mathrm{a}^{\mathrm{5}} \:\:\:\mathrm{c}:\mathrm{3}\left(\mathrm{a}^{\mathrm{10}} +\mathrm{a}^{\mathrm{5}} \right)\:\:\mathrm{d}:\mathrm{0} \\…
Question Number 160058 by Rasheed.Sindhi last updated on 24/Nov/21 $${Find}\:{n}\:{so}\:{that}\:\frac{{a}^{{n}+\mathrm{1}} +{b}^{{n}+\mathrm{1}} }{{a}^{{n}} +{b}^{{n}} }\:{may}\:{be} \\ $$$${the}\:{arithmetic}\:{mean}\:{between}\:{a} \\ $$$${and}\:{b}. \\ $$ Commented by Tinku Tara last…
Question Number 160048 by HongKing last updated on 23/Nov/21 $$\mathrm{Prove}\:\mathrm{that}: \\ $$$$\underset{\:\mathrm{0}} {\overset{\:\mathrm{1}} {\int}}\:\frac{\mathrm{x}\:\mathrm{sin}^{-\mathrm{1}} \:\mathrm{x}}{\mathrm{1}\:+\:\mathrm{sin}^{-\mathrm{1}} \:\mathrm{x}}\:\mathrm{dx}\:<\:\frac{\mathrm{1}}{\mathrm{4}} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
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Question Number 94479 by i jagooll last updated on 19/May/20 $$\mathrm{find}\:\mathrm{solution}\:\mathrm{in}\:\mathbb{C}\: \\ $$$$\mathrm{x}^{\mathrm{4}} +\mathrm{x}^{\mathrm{3}} +\mathrm{x}^{\mathrm{2}} +\mathrm{x}+\mathrm{1}\:=\:\mathrm{0} \\ $$ Answered by john santu last updated on…
Question Number 160008 by HongKing last updated on 23/Nov/21 $$\mathrm{x}_{\mathrm{1}} =\mathrm{3}\:;\:\mathrm{n}\left(\mathrm{x}_{\mathrm{1}} +\mathrm{x}_{\mathrm{2}} +…+\mathrm{x}_{\boldsymbol{\mathrm{n}}} \right)=\mathrm{x}_{\boldsymbol{\mathrm{n}}} \:;\:\mathrm{n}\in\mathbb{N}\:;\:\mathrm{n}\geqslant\mathrm{1} \\ $$$$\mathrm{Find}: \\ $$$$\Omega\:=\underset{\boldsymbol{\mathrm{n}}=\mathrm{1}} {\overset{\infty} {\sum}}\left(-\mathrm{1}\right)^{\boldsymbol{\mathrm{n}}+\mathrm{1}} \:\mathrm{x}_{\boldsymbol{\mathrm{n}}} \\ $$ Commented…
Question Number 160009 by HongKing last updated on 23/Nov/21 $$\mathrm{Find}: \\ $$$$\underset{\boldsymbol{\mathrm{n}}\rightarrow\infty} {\mathrm{lim}}\left(\sqrt[{\boldsymbol{\mathrm{n}}}]{\mathrm{n}!}\:\centerdot\underset{\frac{\mathrm{1}}{\mathrm{1}^{\mathrm{2}} }\:+\:\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{2}} }\:+\:…\:+\:\frac{\mathrm{1}}{\boldsymbol{\mathrm{n}}^{\mathrm{2}} }} {\overset{\:\frac{\boldsymbol{\pi}^{\mathrm{2}} }{\mathrm{6}}} {\int}}\:\mathrm{e}^{\boldsymbol{\mathrm{x}}^{\mathrm{2}} } \:\mathrm{dx}\right) \\ $$ Answered by…
Question Number 160006 by HongKing last updated on 23/Nov/21 $$\mathrm{Evaluate}: \\ $$$$\underset{\boldsymbol{\mathrm{n}}\rightarrow\infty} {\mathrm{lim}}\:\underset{\boldsymbol{\mathrm{n}}} {\overset{\boldsymbol{\mathrm{n}}+\mathrm{1}} {\int}}\:\mathrm{e}^{\frac{\mathrm{1}}{\boldsymbol{\mathrm{x}}}} \:\mathrm{dx}\:=\:? \\ $$$$ \\ $$ Commented by kowalsky78 last updated…