Question Number 197860 by cortano12 last updated on 01/Oct/23 $$\:\:\:\:\frac{\mathrm{2}^{\mathrm{17}} +\mathrm{2}^{\mathrm{16}} +\mathrm{2}^{\mathrm{15}} +\ldots+\mathrm{1}}{\mathrm{2}^{\mathrm{8}} +\mathrm{2}^{\mathrm{7}} +\mathrm{2}^{\mathrm{6}} +\ldots+\mathrm{1}}\:=\:?\: \\ $$ Answered by mr W last updated on…
Question Number 197843 by a.lgnaoui last updated on 30/Sep/23 $$\boldsymbol{\mathrm{Exercice}}\:\:\mathrm{2} \\ $$ Commented by a.lgnaoui last updated on 30/Sep/23 Commented by witcher3 last updated on…
Question Number 197838 by hardmath last updated on 30/Sep/23 Answered by MM42 last updated on 30/Sep/23 $$\frac{\mathrm{2}^{\mathrm{49}} }{\mathrm{2}^{\mathrm{49}} +\mathrm{1}}+\frac{\mathrm{2}^{\mathrm{48}} }{\mathrm{2}^{\mathrm{48}} +\mathrm{1}}+…+\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{48}} +\mathrm{1}}+\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{49}} +\mathrm{1}} \\ $$$$=\mathrm{1}+\mathrm{1}+…+\mathrm{1}=\:\mathrm{49}\:\checkmark…
Question Number 197784 by cortano12 last updated on 28/Sep/23 $$\:\sqrt[{\mathrm{3}}]{\mathrm{x}−\mathrm{2}+\mathrm{3}\sqrt[{\mathrm{3}}]{\mathrm{x}−\mathrm{3}}\:\left(\mathrm{1}+\sqrt[{\mathrm{3}}]{\mathrm{x}−\mathrm{3}}\right)}\:+\:\sqrt[{\mathrm{3}}]{\mathrm{x}+\mathrm{5}+\mathrm{6}\sqrt[{\mathrm{3}}]{\mathrm{x}−\mathrm{3}}\left(\mathrm{1}+\mathrm{2}\sqrt[{\mathrm{3}}]{\mathrm{x}−\mathrm{3}}\:\right)}\:=\:\mathrm{5} \\ $$ Answered by Frix last updated on 28/Sep/23 $$\mathrm{Let}\:{x}={t}^{\mathrm{3}} +\mathrm{3} \\ $$$$\sqrt[{\mathrm{3}}]{\left({t}+\mathrm{1}\right)^{\mathrm{3}} }+\sqrt[{\mathrm{3}}]{{t}^{\mathrm{3}} +\mathrm{12}{t}^{\mathrm{2}}…
Question Number 197794 by universe last updated on 28/Sep/23 $$\:\:\:\:\mathrm{if}\:\mathrm{x}\:\:\:=\:\:\:\mathrm{log}\:\mathrm{tan}\left(\frac{\pi}{\mathrm{4}}+\frac{\mathrm{y}}{\mathrm{2}}\right),\:\:\mathrm{prove}\:\mathrm{that} \\ $$$$\:\:\:\:\:\mathrm{y}\:\:\:\:=\:\:\:−{i}\mathrm{log}\:\mathrm{tan}\left(\frac{{ix}}{\mathrm{2}}\:+\:\frac{\pi}{\mathrm{4}}\right)\:\:\:\:\:\mathrm{here}\:{i}\:\:=\:\sqrt{−\mathrm{1}} \\ $$ Answered by Frix last updated on 29/Sep/23 $${x}=\mathrm{ln}\:\mathrm{tan}\:\frac{\mathrm{2}{y}+\pi}{\mathrm{4}}\:\Leftrightarrow\:{y}=−\frac{\pi}{\mathrm{2}}+\mathrm{2tan}^{−\mathrm{1}} \:\mathrm{e}^{{x}} \\ $$$$−\frac{\pi}{\mathrm{2}}+\mathrm{2tan}^{−\mathrm{1}}…
Question Number 197708 by karowan last updated on 26/Sep/23 $$ \\ $$$${x}^{{a}} ={x}^{{a}+\mathrm{4}} \:\:\mathrm{where}\:{a}\in\mathbb{Z} \\ $$$$\mathrm{solve}\:\mathrm{for}\:{x}\:\mathrm{by}\:\mathrm{showing}\:\mathrm{steps} \\ $$ Answered by Frix last updated on 26/Sep/23…
Question Number 197706 by universe last updated on 26/Sep/23 Answered by witcher3 last updated on 27/Sep/23 $$\left(\frac{\sqrt{\mathrm{a}}+\sqrt{\mathrm{b}}}{\mathrm{2}}\right)^{\mathrm{2}} \leqslant\frac{\mathrm{a}+\mathrm{b}}{\mathrm{2}}\Leftrightarrow\frac{\mathrm{a}+\mathrm{b}}{\mathrm{4}}\geqslant\frac{\sqrt{\mathrm{ab}}}{\mathrm{2}}.\mathrm{Am}−\mathrm{GM} \\ $$$$\left.\Rightarrow\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }}+\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}+\mathrm{y}^{\mathrm{2}} }}\leqslant\mathrm{2}\sqrt{\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{\mathrm{1}}{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{1}+\mathrm{y}^{\mathrm{2}} }\right.}\right) \\…
Question Number 197623 by hardmath last updated on 24/Sep/23 $$\mathrm{x},\mathrm{y}\in\mathbb{N} \\ $$$$\mathrm{162}\:\centerdot\:\mathrm{x}^{\mathrm{2}} \:=\:\mathrm{y}^{\mathrm{3}} \\ $$$$\mathrm{min}\left(\mathrm{x}+\mathrm{y}\right)=? \\ $$ Commented by SANOGO last updated on 25/Sep/23 merci bien monsieur…
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} \\…