Question Number 163588 by HongKing last updated on 08/Jan/22 $${In}\:\:\bigtriangleup{ABC}\:\:{prove}\:{that} \\ $$$$\frac{{a}}{{b}}\:+\:\frac{{b}}{{c}}\:+\:\frac{{c}}{{a}}\:+\:\frac{{R}^{\mathrm{2}} }{\mathrm{4}{r}^{\mathrm{2}} }\:\geqslant\:\mathrm{1}\:+\:\frac{{b}^{\mathrm{2}} }{{a}^{\mathrm{2}} }\:+\:\frac{{c}^{\mathrm{2}} }{{b}^{\mathrm{2}} }\:+\:\frac{{a}^{\mathrm{2}} }{{c}^{\mathrm{2}} } \\ $$ Terms of Service…
Question Number 163587 by HongKing last updated on 08/Jan/22 Answered by MJS_new last updated on 08/Jan/22 $$\int\frac{{x}}{\left({x}+\mathrm{1}\right)\left({x}^{\mathrm{2}} +\mathrm{1}\right)\left({a}^{\mathrm{2}} {x}^{\mathrm{2}} +\mathrm{1}\right)}{dx}= \\ $$$$=−\frac{\mathrm{1}}{\mathrm{2}\left({a}^{\mathrm{2}} +\mathrm{1}\right)}\int\frac{{dx}}{{x}+\mathrm{1}}−\frac{\mathrm{1}}{\mathrm{2}\left({a}^{\mathrm{2}} −\mathrm{1}\right)}\int\frac{{x}+\mathrm{1}}{{x}^{\mathrm{2}} +\mathrm{1}}{dx}+\frac{{a}^{\mathrm{2}}…
Question Number 163586 by HongKing last updated on 08/Jan/22 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 163576 by cortano1 last updated on 08/Jan/22 $$\:\:{What}\:{is}\:{the}\:{coefficient}\:{of}\:{x}^{\mathrm{2020}} \\ $$$$\:{in}\:\left(\mathrm{1}+{x}+{x}^{\mathrm{2}} +{x}^{\mathrm{3}} +…+{x}^{\mathrm{2020}} \right)^{\mathrm{2021}} \\ $$ Answered by bobhans last updated on 08/Jan/22 $$\:\:\:\mathrm{f}\left(\mathrm{x}\right)=\left(\mathrm{1}+\mathrm{x}+\mathrm{x}^{\mathrm{2}}…
Question Number 32494 by Tinkutara last updated on 26/Mar/18 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 163564 by HongKing last updated on 07/Jan/22 $$\mathrm{Find}\:\mathrm{the}\:\mathrm{closed}\:\mathrm{form}\:\mathrm{of}\:\mathrm{a}\:\mathrm{integral}: \\ $$$$\boldsymbol{\Omega}\:\:=\:\underset{\:\frac{\boldsymbol{\pi}}{\mathrm{2}}} {\overset{\:\boldsymbol{\pi}} {\int}}\:\frac{\mathrm{dx}}{\mathrm{e}^{\boldsymbol{\mathrm{x}}} \:+\:\mathrm{ln}\:\left(\mathrm{cos}\boldsymbol{\mathrm{x}}\right)} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 163548 by HongKing last updated on 07/Jan/22 $$\mathrm{Lambert}\:\mathrm{series}\:\mathrm{type}\:\mathrm{representation} \\ $$$$\mathrm{for}\:\:\sqrt{-\mathrm{1}}\:\:\mathrm{factorial} \\ $$$$\left(\mathrm{i}^{\boldsymbol{\mathrm{i}}} \right)!\:\left(-\mathrm{i}^{\boldsymbol{\mathrm{i}}} \right)!\:=\:\mathrm{1}\:-\:\mathrm{2}\:\underset{\boldsymbol{\mathrm{n}}=\mathrm{1}} {\overset{\infty} {\sum}}\:\frac{\left(-\mathrm{1}\right)^{\boldsymbol{\mathrm{n}}} }{\mathrm{e}^{\boldsymbol{\pi}} \:\mathrm{n}^{\mathrm{2}} \:-\:\mathrm{1}} \\ $$ Terms of…
Question Number 163536 by HongKing last updated on 07/Jan/22 $$\mathrm{Prove}\:\mathrm{that}: \\ $$$$\underset{\:\mathrm{9}} {\overset{\:\mathrm{1}} {\int}}\:\frac{\mathrm{ln}\:\left(\mathrm{1}\:+\:\mathrm{x}\right)}{\mathrm{1}\:+\:\mathrm{x}^{\mathrm{2}} }\:\mathrm{dx}\:=\:\frac{\pi\:\centerdot\:\mathrm{ln}\:\left(\mathrm{2}\right)}{\mathrm{8}} \\ $$ Answered by Ar Brandon last updated on 07/Jan/22…
Question Number 32446 by Mr eaay last updated on 25/Mar/18 Answered by MJS last updated on 25/Mar/18 $$\mathrm{left}\:\mathrm{side}: \\ $$$$\mathrm{8}^{\frac{\left({x}−\mathrm{1}\right){x}^{\mathrm{2}{x}} }{{x}^{{x}} \left({x}−\mathrm{2}\right)}} ×\mathrm{16}^{{x}^{{x}} } =\mathrm{8}^{\frac{\left({x}−\mathrm{1}\right){x}^{{x}}…
Question Number 32441 by jasno91 last updated on 25/Mar/18 Answered by MJS last updated on 25/Mar/18 $$\left(\mathrm{1}−\left(\frac{\mathrm{1}}{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{4}}\right)\right)×\mathrm{5}\frac{\mathrm{1}}{\mathrm{2}}{m}= \\ $$$$=\left(\mathrm{1}−\left(\frac{\mathrm{4}}{\mathrm{12}}+\frac{\mathrm{3}}{\mathrm{12}}\right)\right)×\frac{\mathrm{11}}{\mathrm{2}}{m}= \\ $$$$=\left(\frac{\mathrm{12}}{\mathrm{12}}−\frac{\mathrm{7}}{\mathrm{12}}\right)×\frac{\mathrm{11}}{\mathrm{2}}{m}=\frac{\mathrm{5}}{\mathrm{12}}×\frac{\mathrm{11}}{\mathrm{2}}{m}= \\ $$$$=\frac{\mathrm{55}}{\mathrm{24}}{m}=\mathrm{2}\frac{\mathrm{7}}{\mathrm{24}}{m} \\ $$…