Question Number 163906 by HongKing last updated on 11/Jan/22 $$\begin{cases}{\mathrm{x}^{\mathrm{3}} \:+\:\mathrm{x}\:+\:\mathrm{6}\:=\:\mathrm{8y}}\\{\mathrm{y}^{\mathrm{3}} \:+\:\mathrm{y}\:+\:\mathrm{6}\:=\:\mathrm{8z}}\\{\mathrm{z}^{\mathrm{3}} \:+\:\mathrm{z}\:+\:\mathrm{6}\:=\:\mathrm{8x}}\end{cases}\:\:\:\Rightarrow\:\:\:\mathrm{x};\mathrm{y};\mathrm{z}\:=\:? \\ $$ Answered by ajfour last updated on 11/Jan/22 $${let}\:\:{from}\:{symmetry} \\ $$$${x}={y}={z}\:\:\Rightarrow\:\:{x}^{\mathrm{3}}…
Question Number 32832 by Rasheed.Sindhi last updated on 03/Apr/18 $$\mathrm{Determine}\:\mathrm{n}\:\mathrm{such}\:\mathrm{that}\:\mathrm{1001n}+\mathrm{1}\:\mathrm{is} \\ $$$$\mathrm{perfect}\:\mathrm{cube}. \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 163897 by HongKing last updated on 11/Jan/22 $$\mathrm{if}\:\:\mathrm{x}^{\mathrm{3}} \:=\:\mathrm{1}\:\:\mathrm{and}\:\:\mathrm{x}\:\neq\:\mathrm{1} \\ $$$$\mathrm{simplificar}\:\:\left(\frac{\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{4}} }}{\mathrm{1}\:+\:\mathrm{x}^{\mathrm{5}} }\right)^{\mathrm{3}} \\ $$ Answered by mr W last updated on 11/Jan/22…
Question Number 163902 by HongKing last updated on 11/Jan/22 $$\sqrt{\mathrm{x}!^{\boldsymbol{\mathrm{x}}!} }\:\:+\:\:\mathrm{2}^{\boldsymbol{\mathrm{x}}!} \:\:=\:\mathrm{x}!^{\mathrm{3}} \:\:+\:\:\mathrm{10x}!\:\:+\:\:\mathrm{4} \\ $$$$\mathrm{find}:\:\:\boldsymbol{\mathrm{x}}\:=\:? \\ $$ Answered by MJS_new last updated on 11/Jan/22 $${x}!\:\mathrm{is}\:\mathrm{defined}\:\mathrm{for}\:{x}\in\mathbb{N}\:\Rightarrow\:\mathrm{try}\:\mathrm{the}\:\mathrm{first}\:\mathrm{few}……
Question Number 163899 by HongKing last updated on 11/Jan/22 $$\mathrm{Find}:\:\:\boldsymbol{\Omega}\:=\:\underset{\:\mathrm{0}} {\overset{\:\mathrm{1}} {\int}}\:\frac{\mathrm{cos}\left(\mathrm{ax}\right)}{\:\sqrt{\mathrm{x}}\:\centerdot\:\sqrt{\mathrm{1}\:-\:\mathrm{x}}}\:\mathrm{dx} \\ $$ Answered by Kamel last updated on 11/Jan/22 $$\Omega\left({a}\right)=\pi{cos}\left(\frac{{a}}{\mathrm{2}}\right){J}_{\mathrm{0}} \left(\frac{{a}}{\mathrm{2}}\right) \\ $$…
Question Number 163891 by HongKing last updated on 11/Jan/22 $$\mathrm{if}\:\:\:\mathrm{f}\left(\mathrm{x}^{\mathrm{3}} \:+\:\mathrm{1}\right)\:=\:\mathrm{x}^{\mathrm{5}} \:+\:\mathrm{4x}\:+\:\mathrm{2} \\ $$$$\mathrm{find}\:\:\:\underset{\:\mathrm{0}} {\overset{\:\mathrm{1}} {\int}}\:\mathrm{f}\left(\mathrm{x}\right)\:\mathrm{dx}\:=\:? \\ $$ Answered by Ar Brandon last updated on…
Question Number 163865 by mathlove last updated on 11/Jan/22 Answered by TheSupreme last updated on 11/Jan/22 $${x}\neq\pm\mathrm{1} \\ $$$$ \\ $$ Commented by mr W…
Question Number 163861 by mathlove last updated on 11/Jan/22 Answered by Ar Brandon last updated on 11/Jan/22 $${x}=\sqrt{\mathrm{5}}+\frac{\mathrm{1}}{{x}}\Rightarrow{x}=\frac{\sqrt{\mathrm{5}}\pm\sqrt{\mathrm{9}}}{\mathrm{2}}=\frac{\sqrt{\mathrm{5}}+\mathrm{3}}{\mathrm{2}}\:\left({x}>\mathrm{0},\:\sqrt{\mathrm{5}}−\mathrm{3}<\mathrm{0}\right) \\ $$ Terms of Service Privacy Policy…
Question Number 163860 by mathlove last updated on 11/Jan/22 Answered by Ar Brandon last updated on 11/Jan/22 $$\Rightarrow\sqrt[{\mathrm{4}}]{{x}}+\frac{\mathrm{1}}{\:\sqrt[{\mathrm{4}}]{{x}}}+\sqrt[{\mathrm{4}}]{{x}}+\frac{\mathrm{1}}{\:\sqrt[{\mathrm{4}}]{{x}}}=\mathrm{5} \\ $$$$\Rightarrow\mathrm{2}\sqrt[{\mathrm{4}}]{{x}}+\frac{\mathrm{2}}{\:\sqrt[{\mathrm{4}}]{{x}}}=\mathrm{5} \\ $$$$\Rightarrow\mathrm{2}\sqrt[{\mathrm{4}}]{{x}^{\mathrm{2}} }−\mathrm{5}\sqrt[{\mathrm{4}}]{{x}}+\mathrm{2}=\mathrm{0} \\ $$$$\Rightarrow\sqrt[{\mathrm{4}}]{{x}}=\frac{\mathrm{5}\pm\sqrt{\mathrm{9}}}{\mathrm{4}}…
Question Number 32788 by rahul 19 last updated on 02/Apr/18 $$\boldsymbol{{T}}{he}\:{least}\:{positive}\:{integral}\:{value}\:{of} \\ $$$$'{x}'\:{satisfying}\:: \\ $$$$\left({e}^{{x}} −\mathrm{2}\right)\left(\mathrm{sin}\:\left({x}+\frac{\pi}{\mathrm{4}}\right)\right)\left({x}−\mathrm{log}_{{e}} \:\underset{} {\mathrm{2}}\right)\left({sinx}\:−\:{cosx}\right)<\mathrm{0} \\ $$ Commented by rahul 19 last…