Question Number 199709 by tri26112004 last updated on 08/Nov/23 $${u}_{\mathrm{1}} \:=\:\mathrm{2}\: \\ $$$${u}_{{n}+\mathrm{1}} \:=\:\mathrm{3}{u}_{{n}} \:+\:\mathrm{2} \\ $$$${u}_{{n}} \:\rightarrow\:{n}\:¿ \\ $$ Answered by mr W last…
Question Number 199711 by Mingma last updated on 08/Nov/23 Answered by mr W last updated on 08/Nov/23 Commented by mr W last updated on 08/Nov/23…
Question Number 199732 by Rupesh123 last updated on 08/Nov/23 Answered by Rasheed.Sindhi last updated on 08/Nov/23 $${y}=\frac{{x}^{\mathrm{2}} +\mathrm{2}}{{x}+\mathrm{3}}=\frac{{x}^{\mathrm{2}} −\mathrm{9}+\mathrm{9}+\mathrm{2}}{{x}+\mathrm{3}} \\ $$$$\:\:\:=\frac{\left({x}−\mathrm{3}\right)\left({x}+\mathrm{3}\right)+\mathrm{11}}{{x}+\mathrm{3}} \\ $$$$=\left({x}−\mathrm{3}\right)+\frac{\mathrm{11}}{{x}+\mathrm{3}}\in\mathbb{Z}\Rightarrow\left({x}+\mathrm{3}\right)\mid\mathrm{11} \\ $$$${x}+\mathrm{3}=\pm\mathrm{1},\pm\mathrm{11}…
Question Number 199733 by Rupesh123 last updated on 08/Nov/23 Answered by mr W last updated on 08/Nov/23 $${x}^{\mathrm{6}} −{x}^{\mathrm{4}} +\mathrm{1}=\mathrm{0} \\ $$$$\left({x}^{\mathrm{2}} \right)^{\mathrm{3}} −\left({x}^{\mathrm{2}} \right)^{\mathrm{2}}…
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Question Number 199731 by Rupesh123 last updated on 08/Nov/23 Answered by des_ last updated on 08/Nov/23 $$\left({x}\:+\:\frac{\mathrm{1}}{{x}}\:\right)^{\mathrm{3}} =\:\mathrm{3}\sqrt{\mathrm{3}}\:\Rightarrow \\ $$$$\Rightarrow\:{x}^{\mathrm{3}} \:+\:\mathrm{3}\left({x}\:+\:\frac{\mathrm{1}}{{x}}\:\right)\:+\:\frac{\mathrm{1}}{{x}^{\mathrm{3}} }\:=\:\mathrm{3}\sqrt{\mathrm{3}}\:\:\Rightarrow \\ $$$$\Rightarrow\:{x}^{\mathrm{3}} \:+\:\frac{\mathrm{1}}{{x}^{\mathrm{3}}…
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Question Number 199721 by galivan last updated on 08/Nov/23 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 199671 by Mingma last updated on 07/Nov/23 Answered by mr W last updated on 07/Nov/23 $${a}_{\mathrm{1}} =\mathrm{3},\:{a}_{\mathrm{2}} =\mathrm{9}\:\Rightarrow{a}_{\mathrm{0}} =−\mathrm{1} \\ $$$${a}_{{n}} =\frac{{a}_{{n}−\mathrm{1}} +{a}_{{n}+\mathrm{1}}…
Question Number 199694 by Mingma last updated on 07/Nov/23 Answered by witcher3 last updated on 07/Nov/23 $$\mathrm{g}\left(\mathrm{x}\right)=\mathrm{f}\left(\frac{\mathrm{x}+\mathrm{b}}{\mathrm{2}}\right)−\frac{\mathrm{f}\left(\mathrm{x}\right)+\mathrm{f}\left(\mathrm{b}\right)}{\mathrm{2}} \\ $$$$\mathrm{g}\left(\mathrm{a}\right)=\mathrm{0},\mathrm{g}\left(\mathrm{b}\right)=\mathrm{0},\mathrm{rohll}\:\mathrm{theorem} \\ $$$$\left.\Rightarrow\exists\mathrm{c}\in\right]\mathrm{a},\mathrm{b}\left[\mid\mathrm{g}'\left(\mathrm{c}\right)=\mathrm{0};\right. \\ $$$$\mathrm{g}'\left(\mathrm{x}\right)=\frac{\mathrm{1}}{\mathrm{2}}\mathrm{f}'\left(\frac{\mathrm{x}+\mathrm{b}}{\mathrm{2}}\right)−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{f}'\left(\mathrm{x}\right) \\ $$$$\mathrm{g}'\left(\mathrm{c}\right)=\mathrm{0}\Leftrightarrow\mathrm{f}'\left(\frac{\mathrm{c}+\mathrm{b}}{\mathrm{2}}\right)=\mathrm{f}'\left(\mathrm{c}\right)\:\mathrm{applie}\:\mathrm{Rohll}\:\mathrm{to}…