Question Number 148446 by puissant last updated on 28/Jul/21 $$\mathrm{Soit}\:\mathrm{a},\mathrm{b},\mathrm{c}\:\mathrm{et}\:\alpha\:\mathrm{4}\:\mathrm{nombres}\:\mathrm{rationnels} \\ $$$$\mathrm{telque}\:\sqrt[{\mathrm{3}}]{\alpha}\:\mathrm{est}\:\mathrm{irrationnel}.. \\ $$$$\mathrm{Demontrer}\:\mathrm{que}\:: \\ $$$$\left(\mathrm{a}\sqrt[{\mathrm{3}}]{\alpha}+\mathrm{b}\sqrt[{\mathrm{3}}]{\alpha}=\mathrm{c}\right)\:\Rightarrow\:\left(\mathrm{a}=\mathrm{b}=\mathrm{c}\right).. \\ $$ Commented by Olaf_Thorendsen last updated on 28/Jul/21…
Question Number 148441 by qaz last updated on 28/Jul/21 $$\underset{\mathrm{k}=\mathrm{1}} {\overset{\infty} {\sum}}\underset{\mathrm{m}=\mathrm{1}} {\overset{\mathrm{n}} {\sum}}\frac{\mathrm{n}\left(\mathrm{m}−\mathrm{1}\right)}{\left(\mathrm{nk}+\mathrm{m}−\mathrm{1}\right)\left(\mathrm{nk}+\mathrm{m}\right)}=? \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 148443 by peter frank last updated on 28/Jul/21 $${Can}\:{i}\:{use}\:{this}\:{app}\:\:{on}\:{PC} \\ $$$${to}\:{tinkutara} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 148437 by aliibrahim1 last updated on 28/Jul/21 Answered by mathmax by abdo last updated on 28/Jul/21 $$\mathrm{I}=\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{\mathrm{dx}}{\:\sqrt{\mathrm{1}+\mathrm{x}}+\sqrt{\mathrm{1}−\mathrm{x}}}\:\Rightarrow\mathrm{I}=\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{\sqrt{\mathrm{1}+\mathrm{x}}−\sqrt{\mathrm{1}−\mathrm{x}}}{\mathrm{1}+\mathrm{x}−\mathrm{1}+\mathrm{x}}\mathrm{dx} \\ $$$$=\int_{\mathrm{0}}…
Question Number 148439 by liberty last updated on 28/Jul/21 $$\:\:\underset{{x}\rightarrow\mathrm{2}} {\mathrm{lim}}\:\frac{\sqrt{\mathrm{x}+\mathrm{2}}\:\sqrt[{\mathrm{3}}]{\mathrm{x}+\mathrm{6}}−\mathrm{x}^{\mathrm{2}} }{\mathrm{x}−\mathrm{2}}\:=? \\ $$ Answered by dumitrel last updated on 28/Jul/21 $$\underset{{x}\rightarrow\mathrm{2}} {{lim}}\frac{\sqrt{{x}+\mathrm{2}}\left(\sqrt[{\mathrm{3}}]{{x}+\mathrm{6}}−\mathrm{2}\right)}{{x}−\mathrm{2}}+\underset{{x}\rightarrow\mathrm{2}} {{lim}}\frac{\mathrm{2}\left(\sqrt{{x}+\mathrm{2}}−\mathrm{2}\right)}{{x}−\mathrm{2}}−\underset{{x}\rightarrow\mathrm{2}} {{lim}}\frac{{x}^{\mathrm{2}}…
Question Number 82901 by Power last updated on 25/Feb/20 Answered by mr W last updated on 26/Feb/20 Commented by Power last updated on 26/Feb/20 $$\mathrm{sir}\:\:\mathrm{a}:\mathrm{b}:\mathrm{c}=?…
Question Number 82897 by aseer imad last updated on 25/Feb/20 Commented by mr W last updated on 25/Feb/20 $${certainly}! \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}{f}\left({x}\right)={f}\left(\mathrm{0}\right)=\mathrm{0}+\mathrm{1}+\mathrm{2}=\mathrm{3} \\ $$$$\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}{f}\left({x}\right)={f}\left(\mathrm{1}\right)=\mathrm{1}+\mathrm{0}+\mathrm{1}=\mathrm{2}…
Question Number 17359 by ajfour last updated on 04/Jul/17 $$\mathrm{For}\:\mathrm{what}\:\mathrm{values}\:\mathrm{of}\:\mathrm{m},\:\mathrm{the}\:\mathrm{equation} \\ $$$$\left(\mathrm{1}+\mathrm{m}\right)\mathrm{x}^{\mathrm{2}} −\mathrm{2}\left(\mathrm{1}+\mathrm{3m}\right)\mathrm{x}+\left(\mathrm{1}+\mathrm{8m}\right)=\mathrm{0}\:; \\ $$$$\:\:\:\mathrm{m}\:\in\:\mathrm{R}\:,\:\mathrm{has}\:\mathrm{both}\:\mathrm{roots}\:\mathrm{positive}\:? \\ $$ Commented by ajfour last updated on 04/Jul/17 $$\mathrm{book}'\mathrm{s}\:\mathrm{answer}:…
Question Number 148428 by 0731619 last updated on 27/Jul/21 Answered by Mathspace last updated on 27/Jul/21 $$\frac{\left({a}+{b}\right)^{\mathrm{2}} }{{ab}}−\mathrm{3}\:=\frac{\left({a}+{b}\right)^{\mathrm{2}} −\mathrm{3}{ab}}{{ab}} \\ $$$$=\frac{{a}^{\mathrm{2}} +\mathrm{2}{ab}+{b}^{\mathrm{2}} −\mathrm{3}{ab}}{{ab}} \\ $$$$=\frac{{a}^{\mathrm{2}}…
Question Number 82891 by jagoll last updated on 25/Feb/20 $$\int\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}−\mathrm{x}^{\mathrm{4}} }}\:\mathrm{dx}\:=\:? \\ $$ Answered by mind is power last updated on 25/Feb/20 $$\frac{\mathrm{1}}{.\sqrt{\mathrm{1}−{t}^{\mathrm{2}} }}=\underset{{n}\geqslant\:\mathrm{0}} {\sum}\frac{\left(\mathrm{2}{n}\right)!{t}^{\mathrm{2}{n}}…