Question Number 159173 by HongKing last updated on 13/Nov/21 Answered by qaz last updated on 14/Nov/21 $$\Omega=\underset{\mathrm{k}=\mathrm{1}} {\overset{\infty} {\sum}}\mathrm{arctan}\:\left(\frac{\mathrm{1}}{\mathrm{1}+\left(\mathrm{k}+\frac{\mathrm{5}}{\mathrm{3}}\right)\left(\mathrm{k}+\frac{\mathrm{8}}{\mathrm{3}}\right)}\right) \\ $$$$=\underset{\mathrm{k}=\mathrm{1}} {\overset{\infty} {\sum}}\mathrm{arctan}\:\frac{\left(\mathrm{k}+\frac{\mathrm{8}}{\mathrm{3}}\right)−\left(\mathrm{k}+\frac{\mathrm{5}}{\mathrm{3}}\right)}{\mathrm{1}+\left(\mathrm{k}+\frac{\mathrm{5}}{\mathrm{3}}\right)\left(\mathrm{k}+\frac{\mathrm{8}}{\mathrm{3}}\right)} \\ $$$$=\underset{\mathrm{k}=\mathrm{1}}…
Question Number 159170 by mnjuly1970 last updated on 13/Nov/21 Commented by cortano last updated on 13/Nov/21 $$\Rightarrow\sqrt[{\mathrm{3}}]{{x}}\:+\sqrt[{\mathrm{3}}]{{a}−{x}}\:−\sqrt[{\mathrm{3}}]{\mathrm{2}{a}−{x}}\:=\:\mathrm{0} \\ $$$$\Rightarrow{x}+{a}−{x}−\mathrm{2}{a}+{x}\:=\:−\mathrm{3}\sqrt[{\mathrm{3}}]{{x}\left({a}−{x}\right)\left(\mathrm{2}{a}−{x}\right)} \\ $$$$\Rightarrow{x}−{a}\:=\:−\mathrm{3}\sqrt[{\mathrm{3}}]{{x}\left({a}−{x}\right)\left(\mathrm{2}{a}−{x}\right)} \\ $$$$\Rightarrow{x}^{\mathrm{3}} −\mathrm{3}{ax}^{\mathrm{2}} +\mathrm{3}{a}^{\mathrm{2}}…
Question Number 159152 by HongKing last updated on 13/Nov/21 Answered by mr W last updated on 13/Nov/21 $${let}\:{x}={n}+{f}\:{with}\:{n}\in\mathbb{Z},\:\mathrm{0}\leqslant{f}<\mathrm{1} \\ $$$${RHS}={n}+\mathrm{2022}={LHS}\geqslant\mathrm{2021}{n}\: \\ $$$$\Rightarrow{n}\leqslant\frac{\mathrm{2022}}{\mathrm{2020}}\:\Rightarrow{n}\leqslant\mathrm{1}\:\:\:…\left({i}\right) \\ $$$${RHS}={n}+\mathrm{2022}={LHS}<\mathrm{2021}\left({n}+\mathrm{1}\right)\: \\…
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Question Number 159125 by mr W last updated on 13/Nov/21 $${for}\:{a},\:{b},\:{c}\:>\mathrm{0}\:{and}\:{a}+{b}+{c}=\mathrm{2} \\ $$$${find}\:{min}\left(\mathrm{2}{ab}^{\mathrm{2}} +{b}^{\mathrm{3}} {c},\:\mathrm{2}{bc}^{\mathrm{2}} +{c}^{\mathrm{3}} {a},\:\mathrm{2}{ca}^{\mathrm{2}} +{a}^{\mathrm{3}} {b}\right) \\ $$$${or}\:{disprove}\:{that}\:{such}\:{a}\:{minimum} \\ $$$${doesn}'{t}\:{exist}. \\ $$…
Question Number 159119 by HongKing last updated on 13/Nov/21 $$\mathrm{Assume}\:\:\mathrm{x};\mathrm{y};\mathrm{z}>\mathrm{0}\:\:\mathrm{and}\:\:\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} +\mathrm{z}^{\mathrm{2}} =\mathrm{12} \\ $$$$\mathrm{Prove}\:\mathrm{that}:\:\:\underset{\boldsymbol{\mathrm{cycl}}} {\sum}\:\frac{\frac{\mathrm{x}}{\mathrm{y}}\:+\:\mathrm{1}\:+\:\frac{\mathrm{y}}{\mathrm{x}}}{\frac{\mathrm{1}}{\mathrm{x}}\:+\:\frac{\mathrm{1}}{\mathrm{y}}}\:\leqslant\:\mathrm{9} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 159097 by depressiveshrek last updated on 12/Nov/21 $${f}\left({x}\right)={log}_{\mathrm{2}\:} ^{{x}} \left(\frac{{x}^{\mathrm{2}} −\mathrm{1}}{\:\sqrt{{x}+\mathrm{1}}}\right)+\mid{x}\mid^{\mathrm{2}} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 159092 by HongKing last updated on 12/Nov/21 Answered by Rasheed.Sindhi last updated on 13/Nov/21 $$\left({i}\right){a}\left({a}−{b}\right)=\mathrm{2018} \\ $$$$\begin{cases}{{a}=\mathrm{1}\:\wedge\:{a}−{b}=\mathrm{2018}\Rightarrow{b}=−\mathrm{2017}<\mathrm{0}}\\{{a}=\mathrm{2018}\:\wedge\:{a}−{b}=\mathrm{1}\Rightarrow{b}=\mathrm{2017}}\\{{a}=\mathrm{2}\:\wedge\:{a}−{b}=\mathrm{1009}\Rightarrow{b}=−\mathrm{1007}<\mathrm{0}}\\{{a}=\mathrm{1009}\:\wedge\:{a}−{b}=\mathrm{2}\Rightarrow{b}=\mathrm{1007}}\end{cases} \\ $$$${Possible}\:{solutions}\:\left({a},{b}\right)=\left(\mathrm{2018},\mathrm{2017}\right), \\ $$$$\left(\mathrm{1009},\mathrm{1007}\right) \\ $$$$\left({ii}\right)\:{bc}+{ac}−{b}^{\mathrm{2}}…
Question Number 159086 by HongKing last updated on 12/Nov/21 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 159072 by quvonnn last updated on 12/Nov/21 Commented by quvonnn last updated on 12/Nov/21 $$??? \\ $$ Terms of Service Privacy Policy Contact:…