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Category: Algebra

Question-197876

Question Number 197876 by cortano12 last updated on 02/Oct/23 Commented by Frix last updated on 02/Oct/23 $$\mathrm{You}\:\mathrm{can}\:\mathrm{only}\:\mathrm{approximate} \\ $$$${x}\approx.\mathrm{707349599414} \\ $$$${y}\approx.\mathrm{654856853342} \\ $$$$\mathrm{No}\:\mathrm{other}\:\mathrm{solution}\:\in\mathbb{C} \\ $$…

Question-197838

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…

x-2-3-x-3-1-3-1-x-3-1-3-1-3-x-5-6-x-3-1-3-1-2-x-3-1-3-1-3-5-

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}}…

if-x-log-tan-pi-4-y-2-prove-that-y-ilog-tan-ix-2-pi-4-here-i-1-

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-197706

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) \\…