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Category: Algebra

Question-197464

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]…

find-the-sum-1-x-1-2-x-2-1-4-x-4-1-2-n-x-2n-1-

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-197393

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}} }}}= \\…

Question-197419

Question Number 197419 by universe last updated on 16/Sep/23 Answered by cortano12 last updated on 17/Sep/23 $$\:\:\:\:\begin{cases}{\left(\mathrm{2a}\right)^{\mathrm{ln}\:\mathrm{a}} \:=\:\left(\mathrm{bc}\right)^{\mathrm{ln}\:\mathrm{b}} }\\{\mathrm{b}^{\mathrm{ln}\:\mathrm{2}} \:=\:\mathrm{a}^{\mathrm{ln}\:\mathrm{c}} \:}\end{cases} \\ $$$$\:\:\:\:\:\begin{cases}{\mathrm{ln}\:\left(\mathrm{2a}\right).\:\mathrm{ln}\:\left(\mathrm{a}\right)\:=\:\mathrm{ln}\:\left(\mathrm{b}\right).\:\mathrm{ln}\:\left(\mathrm{bc}\right)}\\{\mathrm{ln}\:\left(\mathrm{b}\right).\:\mathrm{ln}\:\left(\mathrm{2}\right)=\:\mathrm{ln}\:\left(\mathrm{c}\right).\:\mathrm{ln}\:\left(\mathrm{a}\right)}\end{cases} \\ $$$$\:\:\Rightarrow\mathrm{ln}\:\mathrm{a}\:\left\{\mathrm{ln}\:\mathrm{2}+\mathrm{ln}\:\mathrm{a}\right\}=\:\mathrm{ln}\:\mathrm{b}\:\left\{\mathrm{ln}\:\mathrm{b}+\mathrm{ln}\:\mathrm{c}\:\right\}…