Question Number 35892 by behi83417@gmail.com last updated on 25/May/18 Commented by tanmay.chaudhury50@gmail.com last updated on 26/May/18 $${i}\:{am}\:{very}\:{near}\:{to}\:{find}\:{the}\:{answer}…{its}\:{a}\:{very}\:{good} \\ $$$${problem}… \\ $$ Commented by Rasheed.Sindhi last…
Question Number 166958 by cortano1 last updated on 03/Mar/22 $$\:\:\mathrm{If}\:\mathrm{polynomial}\:\mathrm{x}^{\mathrm{3}} −\mathrm{9x}^{\mathrm{2}} +\mathrm{11x}−\mathrm{1}=\mathrm{0}\: \\ $$$$\:\mathrm{have}\:\mathrm{the}\:\mathrm{roots}\:\mathrm{are}\:\mathrm{a},\mathrm{b}\:\mathrm{an}\:\mathrm{c}\:. \\ $$$$\:\mathrm{Given}\:\Delta\:=\:\sqrt{\mathrm{a}}\:+\:\sqrt{\mathrm{b}}\:+\:\sqrt{\mathrm{c}}\:\mathrm{then}\: \\ $$$$\:\:\Delta^{\mathrm{4}} −\mathrm{18}\Delta^{\mathrm{2}} −\mathrm{8}\Delta\:=? \\ $$$$ \\ $$ Commented…
Question Number 166954 by cortano1 last updated on 03/Mar/22 $$\:\mathrm{Given}\:\mathrm{x}^{\mathrm{3}} −\mathrm{3x}^{\mathrm{2}} \sqrt{\mathrm{2}}\:+\mathrm{6x}−\mathrm{2}\sqrt{\mathrm{2}}−\mathrm{8}=\mathrm{0} \\ $$$$\:\mathrm{then}\:\mathrm{x}^{\mathrm{5}} −\mathrm{41x}^{\mathrm{2}} +\mathrm{2022}\:=? \\ $$ Answered by som(math1967) last updated on 03/Mar/22…
Question Number 166921 by cortano1 last updated on 02/Mar/22 Commented by MJS_new last updated on 03/Mar/22 $$\left(\mathrm{2}\right) \\ $$$$\sqrt{{x}}+\sqrt{{y}}=\mathrm{2}{u}>\mathrm{0}\:\left[\mathrm{let}\:{u}=\frac{\mathrm{1}}{\mathrm{4}}\left(\sqrt{\mathrm{5}}−\mathrm{1}\right)\right] \\ $$$$\Rightarrow\:\sqrt{{x}}={u}−{v}\mathrm{i}\wedge\sqrt{{y}}={u}+{v}\mathrm{i}\wedge{u},{v}\:\in\mathbb{R} \\ $$$$\:\:\:\:\:\left[\mathrm{we}\:\mathrm{can}\:\mathrm{let}\:{v}\geqslant\mathrm{0}\:\mathrm{because}\:\mathrm{of}\:\mathrm{symmetry}\right] \\ $$$$\Rightarrow…
Question Number 101363 by bemath last updated on 02/Jul/20 $$\mathrm{The}\:\mathrm{solution}\:\mathrm{set}\:\mathrm{of}\:\mathrm{inequality} \\ $$$$\frac{\sqrt{\left(\mathrm{3x}−\mathrm{7}\right)^{\mathrm{2}} }−\mathrm{2}}{\mathrm{x}−\mathrm{3}}\:\leqslant\:\frac{\mathrm{3}−\sqrt{\mathrm{x}^{\mathrm{2}} }}{\mathrm{x}−\mathrm{3}}\:\mathrm{is}\:\_\_ \\ $$$$\left(\mathrm{A}\right)\:\left(−\infty,\:\frac{\mathrm{1}}{\mathrm{2}}\right]\:\:\:\:\:\:\:\left(\mathrm{D}\right)\:\left[\frac{\mathrm{1}}{\mathrm{2}},\:\infty\right) \\ $$$$\left(\mathrm{B}\right)\:\left[\frac{\mathrm{1}}{\mathrm{2}},\mathrm{1}\:\right]\:\:\:\:\:\:\:\:\:\:\:\:\left(\mathrm{E}\right)\:\left(−\infty,\frac{\mathrm{2}}{\mathrm{3}}\right] \\ $$$$\left(\mathrm{C}\right)\:\left(−\infty,\mathrm{1}\:\right]\: \\ $$ Answered by 1549442205…
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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…