Question Number 226577 by mr W last updated on 06/Dec/25 Answered by Ghisom_ last updated on 06/Dec/25 $${x}_{\mathrm{1}} =\alpha+\sqrt{\beta}+\sqrt{\gamma}+\sqrt{\beta\gamma} \\ $$$${x}_{\mathrm{2}} =\alpha−\sqrt{\beta}−\sqrt{\gamma}+\sqrt{\beta\gamma} \\ $$$${x}_{\mathrm{3}} =\alpha−\sqrt{\beta}+\sqrt{\gamma}−\sqrt{\beta\gamma}…
Question Number 226569 by hardmath last updated on 05/Dec/25 $$\mathrm{a}^{\mathrm{4}} \:+\:\mathrm{b}^{\mathrm{4}} \:+\:\mathrm{c}^{\mathrm{4}} \:=\:\mathrm{2d}^{\mathrm{2}} \\ $$$$\mathrm{Prove}\:\mathrm{that}\:\mathrm{the}\:\mathrm{equation}\:\mathrm{has}\:\mathrm{an}\:\mathrm{infinite} \\ $$$$\mathrm{number}\:\mathrm{of}\:\mathrm{natural}\:\mathrm{solutions} \\ $$ Commented by mr W last updated…
Question Number 226558 by hardmath last updated on 04/Dec/25 Answered by mr W last updated on 04/Dec/25 Commented by hardmath last updated on 05/Dec/25 $$\mathrm{thank}\:\mathrm{you}\:\mathrm{my}\:\mathrm{dear}\:\mathrm{professor}\:\mathrm{cool}…
Question Number 226536 by hardmath last updated on 02/Dec/25 $$\mathrm{If}\:\:\:\left(\mathrm{x}+\left(\mathrm{2a}^{\mathrm{2}} +\mathrm{5}\right)\right)\left(\mathrm{x}−\left(\mathrm{2a}^{\mathrm{2}} +\mathrm{7}\right)\right)\:\leqslant\:\mathrm{0} \\ $$$$\:\:\:\:\:\:\:\mathrm{x}\in\left[−\left(\mathrm{a}^{\mathrm{2}} +\mathrm{8a}−\mathrm{10}\right)\:;\:\left(\mathrm{a}^{\mathrm{2}} +\mathrm{9a}−\mathrm{11}\right)\right] \\ $$$$\mathrm{Find}:\:\boldsymbol{\mathrm{a}}\:=\:? \\ $$ Answered by gregori last updated…
Question Number 226533 by mr W last updated on 02/Dec/25 Answered by mr W last updated on 03/Dec/25 $${cube}\:{roots}\:{of}\:{unit}:\:\mathrm{1},\:\omega,\:\omega^{\mathrm{2}} \\ $$$$\mathrm{1}+\omega+\omega^{\mathrm{2}} =\mathrm{0} \\ $$$${e}^{{x}} =\underset{{n}=\mathrm{0}}…
Question Number 226509 by hardmath last updated on 01/Dec/25 $$\mathrm{Find}:\:\:\:\underset{\boldsymbol{\mathrm{n}}=\mathrm{1}} {\overset{\infty} {\sum}}\:\frac{\mathrm{1}}{\mathrm{n}\centerdot\left(\mathrm{2n}\:+\:\mathrm{1}\right)^{\mathrm{2}} }\:=\:? \\ $$ Answered by mr W last updated on 01/Dec/25 $$\frac{\mathrm{1}}{{n}\left(\mathrm{2}{n}+\mathrm{1}\right)^{\mathrm{2}} }=\frac{{A}}{{n}}+\frac{{B}}{\mathrm{2}{n}+\mathrm{1}}+\frac{{C}}{\left(\mathrm{2}{n}+\mathrm{1}\right)^{\mathrm{2}}…
Question Number 226515 by mr W last updated on 01/Dec/25 Commented by Ghisom_ last updated on 02/Dec/25 $${S}_{{k}} =\underset{{n}=\mathrm{1}} {\overset{{k}} {\sum}}\:\frac{{n}}{{n}^{\mathrm{4}} +{n}^{\mathrm{2}} +\mathrm{1}}\:=\frac{{k}\left({k}+\mathrm{1}\right)}{\mathrm{2}\left({k}^{\mathrm{2}} +{k}+\mathrm{1}\right)} \\…
Question Number 226471 by Rojarani last updated on 30/Nov/25 $$\:{If},\:{x}^{\mathrm{2}} +\mathrm{2}{y}^{\mathrm{2}} \infty{xy}\: \\ $$$$\:\:{then}\:{prove}\:{that},\:\mathrm{2}{x}^{\mathrm{2}} +{y}^{\mathrm{2}} \infty{xy} \\ $$ Commented by fantastic2 last updated on 30/Nov/25…
Question Number 226409 by hardmath last updated on 27/Nov/25 Answered by fantastic2 last updated on 27/Nov/25 Commented by fantastic2 last updated on 27/Nov/25 $$\alpha+\mathrm{90}^{\mathrm{0}} +\mathrm{70}^{\mathrm{0}}…
Question Number 226362 by Rojarani last updated on 26/Nov/25 Answered by Frix last updated on 26/Nov/25 $$\alpha=\sqrt{\mathrm{3}}\mathrm{e}^{−\mathrm{i}\frac{\mathrm{3}\pi}{\mathrm{4}}} \\ $$$$\beta=\sqrt{\mathrm{3}}\mathrm{e}^{\mathrm{i}\frac{\mathrm{3}\pi}{\mathrm{4}}} \\ $$$$\alpha^{{n}} +\beta^{{n}} =\mathrm{2}×\mathrm{3}^{\frac{{n}}{\mathrm{2}}} \mathrm{cos}\:\frac{\mathrm{3}{n}\pi}{\mathrm{4}} \\…