Question Number 219548 by universe last updated on 28/Apr/25 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 219564 by hardmath last updated on 28/Apr/25 $$\mathrm{solve}\:\mathrm{for}\:\mathrm{real}\:\mathrm{numbers}: \\ $$$$\begin{cases}{\frac{\mathrm{xy}\left(\mathrm{x}^{\mathrm{3}} −\mathrm{y}^{\mathrm{3}} \right)+\mathrm{yz}\left(\mathrm{y}^{\mathrm{3}} −\mathrm{z}^{\mathrm{3}} \right)+\mathrm{zx}\left(\mathrm{z}^{\mathrm{3}} −\mathrm{x}^{\mathrm{3}} \right)}{\mathrm{xy}\left(\mathrm{x}−\mathrm{y}\right)+\mathrm{yz}\left(\mathrm{y}−\mathrm{z}\right)+\mathrm{zx}\left(\mathrm{z}−\mathrm{x}\right)}\:=\:\mathrm{55}}\\{\mathrm{x}^{\mathrm{3}} \:+\:\mathrm{y}^{\mathrm{3}} \:+\:\mathrm{z}^{\mathrm{3}} \:=\:\mathrm{99}}\\{\frac{\mathrm{xy}\left(\mathrm{x}^{\mathrm{3}} −\mathrm{y}^{\mathrm{3}} \right)+\mathrm{yz}\left(\mathrm{y}^{\mathrm{3}} −\mathrm{z}^{\mathrm{3}} \right)+\mathrm{zx}\left(\mathrm{z}^{\mathrm{3}}…
Question Number 219561 by hardmath last updated on 28/Apr/25 $$\mathrm{let}\:\mathrm{be}\:\mathrm{the}\:\mathrm{sequence}\:\:\:\left(\mathrm{x}_{\boldsymbol{\mathrm{n}}} \right)\mathrm{n}\:\geqslant\:\mathrm{1} \\ $$$$\mathrm{defined}\:\mathrm{by}\:\:\:\mathrm{x}_{\mathrm{1}} =\:\mathrm{1} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{x}_{\boldsymbol{\mathrm{n}}+\mathrm{2}} \:=\:\mathrm{3x}_{\boldsymbol{\mathrm{n}}+\mathrm{1}} −\:\mathrm{x}_{\boldsymbol{\mathrm{n}}} \\ $$$$\forall\mathrm{n}\:\in\:\mathbb{N} \\ $$$$\mathrm{find}\:\:\:\boldsymbol{\mathrm{L}}\:=\underset{\boldsymbol{\mathrm{n}}\rightarrow\infty} {\mathrm{lim}}\:\sqrt[{\boldsymbol{\mathrm{n}}}]{\underset{\boldsymbol{\mathrm{k}}=\mathrm{0}} {\overset{\mathrm{0}} {\sum}}\:\:\frac{\mathrm{x}_{\mathrm{2}\boldsymbol{\mathrm{k}}+\mathrm{1}}…
Question Number 219562 by hardmath last updated on 28/Apr/25 $$\mathrm{prove}\:\mathrm{that}\:\mathrm{exists}\:\:\:\mathrm{X}\:\in\:\mathrm{M}_{\mathrm{2},\mathrm{3}} \:\left(\mathbb{R}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{Y}\:\in\:\mathrm{M}_{\mathrm{3},\mathrm{2}} \:\left(\mathbb{R}\right) \\ $$$$\mathrm{such}\:\mathrm{that}\:\:\:\mathrm{X}\centerdot\mathrm{Y}\:=\:\begin{pmatrix}{\mathrm{1}}&{\mathrm{1}}\\{\mathrm{1}}&{\mathrm{1}}\end{pmatrix}\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{Y}\centerdot\mathrm{X}\:=\:\begin{pmatrix}{\mathrm{2}}&{\mathrm{6}}&{\mathrm{6}}\\{\mathrm{3}}&{\mathrm{9}}&{\mathrm{9}}\\{-\mathrm{3}}&{-\mathrm{9}}&{-\mathrm{9}}\end{pmatrix}\: \\ $$ Terms of Service Privacy Policy…
Question Number 219563 by hardmath last updated on 28/Apr/25 $$\mathrm{find}\:\mathrm{all}\:\:\:\mathrm{n}\:\in\:\mathbb{N}^{\ast} \\ $$$$\mathrm{such}\:\mathrm{that}\:\:\:\int_{\mathrm{0}} ^{\:\mathrm{1}} \:\left(\mathrm{sinx}\right)^{\mathrm{2n}−\mathrm{2}} \:\centerdot\:\left(\mathrm{cosx}\right)^{\mathrm{2}\boldsymbol{\mathrm{n}}} \:\mathrm{dx}\:\geqslant\:\frac{\mathrm{1}}{\mathrm{4}^{\mathrm{1011}} } \\ $$ Answered by Nicholas666 last updated on…
Question Number 219515 by hardmath last updated on 27/Apr/25 $$\mathrm{Find}:\:\:\:\boldsymbol{\Omega}\:=\:\int_{\mathrm{0}} ^{\:\mathrm{1}} \:\mathrm{x}^{\boldsymbol{\mathrm{x}}^{\mathrm{2}} } \:\mathrm{dx}\:=\:?\: \\ $$ Answered by SdC355 last updated on 27/Apr/25 $$\mathrm{can}'\mathrm{t}\:\mathrm{Find}\:\mathrm{primitive}\:\mathrm{function}\:\int\:\centerdot\: \\…
Question Number 219503 by Rojarani last updated on 27/Apr/25 Answered by Frix last updated on 27/Apr/25 $$\sqrt{\mathrm{4}{x}}+\sqrt{\mathrm{4}{x}−\mathrm{4}}+\sqrt{\mathrm{4}{x}−\mathrm{8}}=\sqrt{{x}+\mathrm{1}}+\sqrt{{x}+\mathrm{5}}+\sqrt{{x}+\mathrm{9}} \\ $$$$\sqrt{\mathrm{4}{x}}=\sqrt{{x}+\mathrm{9}}\:\Rightarrow\:{x}=\mathrm{3} \\ $$$$\sqrt{\mathrm{4}{x}−\mathrm{4}}=\sqrt{{x}+\mathrm{5}}\:\Rightarrow\:{x}=\mathrm{3} \\ $$$$\sqrt{\mathrm{4}{x}−\mathrm{8}}=\sqrt{{x}+\mathrm{1}}\:\Rightarrow\:{x}=\mathrm{3} \\ $$…
Question Number 219476 by universe last updated on 26/Apr/25 $$\:\mathrm{let}\:\mathrm{a}_{\mathrm{1}} \:=\:\mathrm{1}\:;\:\left(\mathrm{n}+\mathrm{1}\right)\mathrm{a}_{\mathrm{n}+\mathrm{1}} +\mathrm{na}_{\mathrm{n}} \:=\:\mathrm{2n}−\mathrm{3}\: \\ $$$$\:\:\mathrm{find}\:\:\mathrm{nth}\:\mathrm{term}\:\mathrm{of}\:\mathrm{a}_{\mathrm{n}\:} \\ $$ Commented by universe last updated on 26/Apr/25 $${yes}\:{sir}\:…
Question Number 219492 by hardmath last updated on 26/Apr/25 $$\mathrm{a}\:+\:\mathrm{b}\:=\:\mathrm{1} \\ $$$$\mathrm{a}^{\mathrm{2}} \:+\:\mathrm{b}^{\mathrm{2}} \:=\:\mathrm{2} \\ $$$$\mathrm{Find}:\:\:\:\mathrm{a}^{\mathrm{11}} \:+\:\mathrm{b}^{\mathrm{11}} \:=\:? \\ $$ Answered by mr W last…
Question Number 219454 by mr W last updated on 25/Apr/25 $${a},\:{b},\:{c}\:{are}\:{the}\:{roots}\:{of}\:{the}\:{equation} \\ $$$${x}^{\mathrm{3}} −\mathrm{3}{x}+\mathrm{1}=\mathrm{0}. \\ $$$${find}\:\sqrt[{\mathrm{3}}]{{a}}+\sqrt[{\mathrm{3}}]{{b}}+\sqrt[{\mathrm{3}}]{{c}}=?\:\&\:\frac{\mathrm{1}}{\:\sqrt[{\mathrm{3}}]{{a}}}+\frac{\mathrm{1}}{\:\sqrt[{\mathrm{3}}]{{b}}}+\frac{\mathrm{1}}{\:\sqrt[{\mathrm{3}}]{{c}}}=? \\ $$ Commented by mr W last updated on…