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Question Number 166830 by HongKing last updated on 28/Feb/22 Answered by nurtani last updated on 01/Mar/22 $$\left({x}+\frac{\mathrm{1}}{{y}}\right)\left({y}+\frac{\mathrm{1}}{{z}}\right)\left({z}+\frac{\mathrm{1}}{{x}}\right)=\left({xy}+\frac{{x}}{{z}}+\mathrm{1}+\frac{\mathrm{1}}{{yz}}\right)\left({z}+\frac{\mathrm{1}}{{x}}\right)={xyz}+{x}+{z}+\frac{\mathrm{1}}{{y}}+{y}+\frac{\mathrm{1}}{{z}}+\frac{\mathrm{1}}{{x}}+\frac{\mathrm{1}}{{xyz}}=\:\left(\mathrm{4}\right)\left(\mathrm{1}\right)\left(\frac{\mathrm{7}}{\mathrm{3}}\right) \\ $$$$\Leftrightarrow\:{xyz}+\frac{\mathrm{1}}{{xyz}}+\left({x}+\frac{\mathrm{1}}{{y}}\right)+\left({y}+\frac{\mathrm{1}}{{z}}\right)+\left({z}+\frac{\mathrm{1}}{{x}}\right)=\left(\mathrm{4}\right)\left(\mathrm{1}\right)\left(\frac{\mathrm{7}}{\mathrm{3}}\right)=\frac{\mathrm{28}}{\mathrm{3}} \\ $$$$\Leftrightarrow\:{xyz}+\frac{\mathrm{1}}{{xyz}}+\mathrm{4}+\mathrm{1}+\frac{\mathrm{7}}{\mathrm{3}}=\frac{\mathrm{28}}{\mathrm{3}} \\ $$$$\Leftrightarrow\:{xyz}+\frac{\mathrm{1}}{{xyz}}+\frac{\mathrm{22}}{\mathrm{3}}=\frac{\mathrm{28}}{\mathrm{3}}\Leftrightarrow\:{xyz}+\frac{\mathrm{1}}{{xyz}}=\frac{\mathrm{6}}{\mathrm{3}}=\mathrm{2} \\ $$$$\Leftrightarrow\:{x}^{\mathrm{2}}…
Question Number 35723 by math2018 last updated on 22/May/18 $${Find}\:{the}\:{largest}\:{prime}\:{factor}\:{of}\:\:{the}\:{following}: \\ $$$$\left(\mathrm{1}×\mathrm{2}×\mathrm{3}\right)+\left(\mathrm{2}×\mathrm{3}×\mathrm{4}\right)+…+\left(\mathrm{2014}×\mathrm{2015}×\mathrm{2016}\right) \\ $$ Answered by MJS last updated on 23/May/18 $$\underset{{i}=\mathrm{1}} {\overset{{n}} {\sum}}\left({i}+\mathrm{2}\right)\left({i}+\mathrm{1}\right){i}=\frac{\mathrm{1}}{\mathrm{4}}\left({n}+\mathrm{3}\right)\left({n}+\mathrm{2}\right)\left({n}+\mathrm{1}\right){n} \\…
Question Number 101225 by harckinwunmy last updated on 01/Jul/20 Commented by Dwaipayan Shikari last updated on 01/Jul/20 $${x}^{{a}} ={y}^{{b}} ={z}^{{c}} ={k}\left({k}\neq\mathrm{0}\right) \\ $$$${x}={k}^{\frac{\mathrm{1}}{{a}}} \:{y}={k}^{\frac{\mathrm{1}}{{b}}} \:\:\:{z}={k}^{\frac{\mathrm{1}}{{c}}}…
Question Number 166756 by HongKing last updated on 27/Feb/22 $$\mathrm{Evaluate}: \\ $$$$\int_{\mathrm{0}} ^{\:\infty} \:\frac{\left(\mathrm{x}\:+\:\mathrm{1}\right)^{\mathrm{2020}} }{\left(\mathrm{x}\:+\:\mathrm{2}\right)^{\mathrm{2022}} }\:\mathrm{dx}\:=\:? \\ $$ Answered by Mathspace last updated on 27/Feb/22…
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Question Number 166754 by mathlove last updated on 27/Feb/22 Answered by mindispower last updated on 28/Feb/22 $${S}=\underset{{n}=\mathrm{1}} {\overset{{m}} {\sum}}{cot}^{\mathrm{2}} \left(\frac{{n}\pi}{\mathrm{2}{m}+\mathrm{1}}\right),\underset{\mathrm{1}} {\overset{\mathrm{2}{m}} {\sum}}{cot}^{\mathrm{2}} \left(\frac{{n}\pi}{\mathrm{2}{m}+\mathrm{1}}\right)=\mathrm{2}{S} \\ $$$${put}\:{z}_{{k}}…
Question Number 35626 by abdo mathsup 649 cc last updated on 21/May/18 $$\left.\mathrm{1}\right)\:{study}\:{the}\:{diagonalisstion}\:{of}\:{the}\:{matrice} \\ $$$${A}\:=\begin{pmatrix}{\mathrm{1}+{a}^{\mathrm{2}} \:\:\:\:\:{a}\:\:\:\:\:\:\:\mathrm{0}}\\{{a}\:\:\:\:\:\:\:\:\:\mathrm{1}+{a}^{\mathrm{2}} \:\:\:\:\:{a}}\end{pmatrix} \\ $$$$\:\:\:\:\:\:\:\:\:\:\left(\:\mathrm{0}\:\:\:\:\:\:\:\:\:\:{a}\:\:\:\:\mathrm{1}+{a}^{\mathrm{2}} \:\right) \\ $$$$\left.\mathrm{2}\right)\:{calculate}\:{A}^{{n}} \\ $$ Terms…
Question Number 35622 by abdo mathsup 649 cc last updated on 21/May/18 $${find}\:{all}\:{matrices}\:{M}\:\in{M}_{\mathrm{3}} \left({R}\right)\:\:/\:\:{M}^{\mathrm{2}} \:={M} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 101156 by I want to learn more last updated on 30/Jun/20 $$\mathrm{Find}\:\mathrm{the}\:\mathrm{range}\:\mathrm{of}\:\mathrm{the}\:\mathrm{function}: \\ $$$$\:\:\:\:\:\mathrm{h}\::\:\mathrm{x}\:\:\:=\:\:\mathrm{2}\:\:−\:\:\mathrm{x}^{\mathrm{2}} \:\mathrm{sin}\left(\mathrm{x}\right),\:\:\:\:\:\:\mathrm{x}\:\:\geqslant\:\:\mathrm{0} \\ $$ Answered by 1549442205 last updated on…