Question Number 197618 by srijanGuha last updated on 24/Sep/23 Answered by Rasheed.Sindhi last updated on 24/Sep/23 $$\mathrm{3}\left({x}+\frac{\mathrm{1}}{{x}}\right)=\mathrm{2} \\ $$$${x}+\frac{\mathrm{1}}{{x}}=\frac{\mathrm{2}}{\mathrm{3}} \\ $$$$\left({x}+\frac{\mathrm{1}}{{x}}\right)^{\mathrm{3}} =\left(\frac{\mathrm{2}}{\mathrm{3}}\right)^{\mathrm{3}} \\ $$$${x}^{\mathrm{3}} +\frac{\mathrm{1}}{{x}^{\mathrm{3}}…
Question Number 197609 by dimentri last updated on 24/Sep/23 $$\:\:\:\frac{{x}+\mathrm{3}}{\mathrm{2022}}\:+\:\frac{{x}+\mathrm{2}}{\mathrm{2023}}\:+\:\frac{{x}+\mathrm{1}}{\mathrm{2024}}\:+\:\frac{{x}}{\mathrm{2025}}\:=\:−\mathrm{4} \\ $$ Answered by Rasheed.Sindhi last updated on 24/Sep/23 $$\:\:\:\frac{{x}+\mathrm{3}}{\mathrm{2022}}\:+\:\frac{{x}+\mathrm{2}}{\mathrm{2023}}\:+\:\frac{{x}+\mathrm{1}}{\mathrm{2024}}\:+\:\frac{{x}}{\mathrm{2025}}\:=\:−\mathrm{4} \\ $$$$\:\:\:\frac{{x}+\mathrm{3}}{\mathrm{2022}}+\mathrm{1}\:+\:\frac{{x}+\mathrm{2}}{\mathrm{2023}}+\mathrm{1}\:+\:\frac{{x}+\mathrm{1}}{\mathrm{2024}}+\mathrm{1}\:+\:\frac{{x}}{\mathrm{2025}}+\mathrm{1}\:=\:−\mathrm{4}+\mathrm{4} \\ $$$$\:\:\:\frac{{x}+\mathrm{2025}}{\mathrm{2022}}\:+\:\frac{{x}+\mathrm{2025}}{\mathrm{2023}}\:+\:\frac{{x}+\mathrm{2025}}{\mathrm{2024}}\:+\:\frac{{x}+\mathrm{2025}}{\mathrm{2025}}\:=\:\mathrm{0} \\…
Question Number 197567 by hardmath last updated on 21/Sep/23 Answered by Frix last updated on 22/Sep/23 $$\mathrm{Obviously}\:{x}={y}=\mathrm{1} \\ $$ Commented by Frix last updated on…
Question Number 197530 by Erico last updated on 20/Sep/23 $$\:\:\:\:\:\mathrm{f}\left(\mathrm{x}\right)−\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{2}}\mathrm{f}''\left(\mathrm{x}\right)=\mathrm{0} \\ $$$$\:\:\:\:\:\:\mathrm{f}\left(\mathrm{x}\right)=? \\ $$ Commented by AST last updated on 20/Sep/23 $${f}\left({x}\right)={x}^{\mathrm{2}} \:{works}. \\…
Question Number 197476 by universe last updated on 19/Sep/23 $$ \\ $$$$\:\:\:\mathrm{find}\:\mathrm{the}\:\mathrm{sum} \\ $$$$\:\frac{\mathrm{1}}{{x}+\mathrm{1}}+\frac{\mathrm{2}}{{x}^{\mathrm{2}} +\mathrm{1}}+\frac{\mathrm{4}}{{x}^{\mathrm{4}} +\mathrm{1}}+………+\frac{\mathrm{2}^{{n}} }{{x}^{\mathrm{2}^{{n}} } +\mathrm{1}}\:\:=\:?? \\ $$ Commented by witcher3 last…
Question Number 197455 by mathlove last updated on 18/Sep/23 $${f}\left({x}\right)=\frac{\mathrm{2}{x}+\mathrm{6}}{\mathrm{2}}+\mathrm{6}−{x} \\ $$$${g}\left({x}\right)=\sqrt{{x}^{\mathrm{99}} +{x}^{\mathrm{98}} +{x}^{\mathrm{97}} +…..+{x}} \\ $$$${f}\left({g}\left(\mathrm{1}\right)\right)+{f}\left({g}\left(\mathrm{2}\right)\right)+………+{f}\left({g}\left(\mathrm{100}\right)\right)=? \\ $$ Answered by hmr last updated on…
Question Number 197464 by universe last updated on 18/Sep/23 Answered by witcher3 last updated on 19/Sep/23 $$\mathrm{claim} \\ $$$$\frac{\left(\mathrm{a}+\mathrm{b}\right)^{\mathrm{6}} }{\left(\mathrm{ab}\right)^{\mathrm{2}} }\geqslant\mathrm{32}\left(\mathrm{a}^{\mathrm{2}} +\mathrm{b}^{\mathrm{2}} \right)….\mathrm{P},\mathrm{by}\:\mathrm{symetri}\:\mathrm{a}\geqslant\mathrm{b} \\ $$$$\left.\mathrm{a}\left.=\mathrm{tb},\mathrm{t}\in\right]\mathrm{0},\mathrm{1}\right]…
Question Number 197465 by universe last updated on 18/Sep/23 $$\:\:\:\mathrm{find}\:\mathrm{the}\:\mathrm{sum} \\ $$$$\:\frac{\mathrm{1}}{{x}+\mathrm{1}}+\frac{\mathrm{2}}{{x}^{\mathrm{2}} +\mathrm{1}}+\frac{\mathrm{4}}{{x}^{\mathrm{4}} +\mathrm{1}}+………+\frac{\mathrm{2}^{{n}} }{{x}^{\mathrm{2}{n}} +\mathrm{1}}\:\:=\:?? \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 197422 by hardmath last updated on 17/Sep/23 Answered by mr W last updated on 23/Sep/23 Commented by mr W last updated on 23/Sep/23…
Question Number 197393 by mathlove last updated on 16/Sep/23 Answered by Frix last updated on 16/Sep/23 $$\frac{\mathrm{3}×\mathrm{5}\sqrt[{\mathrm{3}}]{\mathrm{2}^{\mathrm{2}} \sqrt[{\mathrm{3}}]{\mathrm{2}^{\mathrm{6}} \mathrm{3}}}+\mathrm{3}×\mathrm{7}\sqrt[{\mathrm{3}}]{\mathrm{2}×\mathrm{3}^{\mathrm{2}} \sqrt[{\mathrm{3}}]{\mathrm{3}^{\mathrm{4}} }}}{\:\sqrt[{\mathrm{3}}]{\mathrm{2}^{\mathrm{2}} \mathrm{3}\sqrt[{\mathrm{3}}]{\mathrm{2}^{\mathrm{3}} \mathrm{3}}+\mathrm{2}×\mathrm{3}\sqrt[{\mathrm{3}}]{\mathrm{3}×\mathrm{5}^{\mathrm{3}} }}}= \\…