Question Number 196111 by sonukgindia last updated on 18/Aug/23 Answered by mahdipoor last updated on 18/Aug/23 $$\Rightarrow\Rightarrow\Rightarrow\Rightarrow\:{if}\:{n}\leqslant{x}\:\Rightarrow \\ $$$$\Sigma=\left({x}−\mathrm{1}\right)+…+\left({x}−{n}\right)={nx}−\frac{{n}^{\mathrm{2}} +{n}}{\mathrm{2}}= \\ $$$${x}^{\mathrm{2}} −\left({n}+\mathrm{1}\right){x}+\left(\frac{{n}^{\mathrm{2}} +\mathrm{2}{n}+\mathrm{5}}{\mathrm{4}}\right) \\…
Question Number 196103 by MrGHK last updated on 18/Aug/23 $$\frac{\boldsymbol{\mathrm{d}}^{\boldsymbol{\mathrm{n}}} }{\boldsymbol{\mathrm{dx}}^{\boldsymbol{\mathrm{n}}} }\left[\boldsymbol{\Gamma}\left(\boldsymbol{\mathrm{x}}\right)\right]\:\:\boldsymbol{\mathrm{Hence}}\:\boldsymbol{\mathrm{what}}\:\boldsymbol{\mathrm{is}}\:\frac{\boldsymbol{\mathrm{d}}^{\frac{\mathrm{1}}{\mathrm{2}}} }{\boldsymbol{\mathrm{dx}}^{\frac{\mathrm{1}}{\mathrm{2}}} }\left[\boldsymbol{\Gamma}\left(\boldsymbol{\mathrm{x}}\right)\right] \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 196097 by leicianocosta last updated on 18/Aug/23 Terms of Service Privacy Policy Contact: info@tinkutara.com
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Question Number 196098 by sonukgindia last updated on 18/Aug/23 Answered by mr W last updated on 18/Aug/23 $${x}^{\frac{\mathrm{1}}{\mathrm{2}^{{x}^{\mathrm{4}} } }} ={x}^{\frac{\mathrm{1}}{\mathrm{4}^{{x}^{\mathrm{2}} } }} \\ $$$$\Rightarrow{x}=\mathrm{0}…
Question Number 196063 by Rodier97 last updated on 17/Aug/23 $$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$\mathrm{calcul}\:\:\:\:\int_{\mathrm{0}} ^{+\infty} \:\frac{\sqrt{\mathrm{u}}\:.\mathrm{arctan}\left(\mathrm{u}\right)}{\mathrm{1}+\mathrm{u}^{\mathrm{2}} }\:\mathrm{du} \\ $$$$ \\ $$$$…
Question Number 196062 by qaz last updated on 17/Aug/23 $$\underset{{i}=\mathrm{1}} {\overset{{n}} {\sum}}\underset{{j}=\mathrm{1}} {\overset{{n}} {\sum}}\left[{gcd}\left({i},{j}\right)=\mathrm{1}\right]=?\:\:\:\:\:\:\:\:\:,\left[{D}\right]=\begin{cases}{\mathrm{1},\:\:\:{D}\:{is}\:{ture}.}\\{\mathrm{0},\:\:\:\:{D}\:{is}\:{false}.\:\:\:}\end{cases} \\ $$ Commented by mr W last updated on 17/Aug/23 $${i}\:{don}'{t}\:{think}\:{we}\:{can}\:{get}\:{it}\:{in}\:{a}\:{closed}…
Question Number 196085 by sonukgindia last updated on 17/Aug/23 Answered by MM42 last updated on 17/Aug/23 $$\left(\mathrm{1}^{\mathrm{17}} +\mathrm{2023}^{\mathrm{17}} \right)+\left(\mathrm{2}^{\mathrm{17}} +\mathrm{2022}^{\mathrm{17}} \right)+…+\mathrm{1012}^{\mathrm{17}} = \\ $$$$\left(\mathrm{1}+\mathrm{2023}\right){k}_{\mathrm{1}} +\left(\mathrm{2}+\mathrm{2022}\right){k}_{\mathrm{2}}…
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Question Number 196049 by CrispyXYZ last updated on 17/Aug/23 $${a},\:{b},\:{c}\:>\:\mathrm{0}.\:\mathrm{Find}\:\mathrm{the}\:\mathrm{min}\:\mathrm{value}\:\mathrm{of} \\ $$$$\underset{\mathrm{cyc}} {\sum}\:\sqrt{\frac{{a}+{b}}{{a}+{b}+{c}}}\:. \\ $$ Commented by Frix last updated on 17/Aug/23 $${b}={pa}\wedge{c}={qa} \\ $$$$\Sigma=\frac{\sqrt{{p}+\mathrm{1}}+\sqrt{{p}+{q}}+\sqrt{{q}+\mathrm{1}}}{\:\sqrt{{p}+{q}+\mathrm{1}}}…