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2025-2025-x-mod-17-

Question Number 201418 by cortano12 last updated on 06/Dec/23 $$\:\:\:\:\:\:\mathrm{2025}^{\mathrm{2025}} \:=\:\mathrm{x}\:\left(\mathrm{mod}\:\mathrm{17}\:\right) \\ $$ Answered by mr W last updated on 06/Dec/23 $$\mathrm{2025}^{\mathrm{2025}} \:\left({mod}\:\mathrm{17}\right) \\ $$$$=\left(\mathrm{119}×\mathrm{17}+\mathrm{2}\right)^{\mathrm{2025}}…

0-1-Li-3-x-2-1-x-dx-

Question Number 201473 by MrGHK last updated on 06/Dec/23 $$\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\boldsymbol{\mathrm{Li}}_{\mathrm{3}} \left(−\boldsymbol{\mathrm{x}}^{\mathrm{2}} \right)}{\mathrm{1}+\boldsymbol{\mathrm{x}}}\boldsymbol{\mathrm{dx}} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com

Let-f-x-and-g-x-be-given-by-f-x-1-x-1-x-2-1-x-4-1-x-2018-and-g-x-1-x-1-1-x-3-1-x-5-1-x-2017-Prove-that-f-x-g-x-gt-2-for-any-non-in

Question Number 201441 by dimentri last updated on 06/Dec/23 $${Let}\:{f}\left({x}\right)\:{and}\:{g}\left({x}\right)\:{be}\:{given}\:{by}\: \\ $$$$\:{f}\left({x}\right)=\:\frac{\mathrm{1}}{{x}}\:+\frac{\mathrm{1}}{{x}−\mathrm{2}}\:+\frac{\mathrm{1}}{{x}−\mathrm{4}}\:+\:…\:+\frac{\mathrm{1}}{{x}−\mathrm{2018}} \\ $$$$\:{and}\: \\ $$$$\:\:{g}\left({x}\right)=\frac{\mathrm{1}}{{x}−\mathrm{1}}\:+\frac{\mathrm{1}}{{x}−\mathrm{3}}\:+\frac{\mathrm{1}}{{x}−\mathrm{5}}\:+…+\:\frac{\mathrm{1}}{{x}−\mathrm{2017}}. \\ $$$$\:\:{Prove}\:{that}\:\:\mid\:{f}\left({x}\right)−{g}\left({x}\right)\mid\:>\mathrm{2} \\ $$$$\:\:{for}\:{any}\:{non}−{integer}\:{real}\:{number} \\ $$$$\:\:{x}\:{satisfying}\:\mathrm{0}<{x}<\mathrm{2018}.\: \\ $$ Answered…

Question-201443

Question Number 201443 by sonukgindia last updated on 06/Dec/23 Commented by mr W last updated on 06/Dec/23 $${i}\:{and}\:{many}\:{others}\:{do}\:{our}\:{best}\:{to} \\ $$$${answer}\:{your}\:{questions}.\:{but}\:{why}\: \\ $$$${don}'{t}\:{you}\:{answer}\:{any}\:{question}\:{of}\: \\ $$$${mine}\:{and}\:{others}? \\…

Question-201343

Question Number 201343 by gabi last updated on 05/Dec/23 Answered by aleks041103 last updated on 05/Dec/23 $${A}^{\mathrm{3}} +{A}^{\mathrm{2}} +{A}=\mathrm{0} \\ $$$$\lambda\:{is}\:{eigenvalue} \\ $$$$\Rightarrow\lambda^{\mathrm{3}} +\lambda^{\mathrm{2}} +\lambda=\lambda\left(\lambda^{\mathrm{2}}…