Question Number 220519 by hardmath last updated on 14/May/25 $$\mathrm{Solve}\:\mathrm{in}\:\:\:\mathbb{R} \\ $$$$\frac{\mathrm{15}}{\mathrm{x}^{\mathrm{2}} \:-\:\mathrm{3x}\:+\:\mathrm{4}}\:\:+\:\:\frac{\mathrm{7}}{\mathrm{x}^{\mathrm{2}} \:+\:\mathrm{7x}}\:\:+\:\:\frac{\mathrm{10}}{\mathrm{x}^{\mathrm{2}} \:+\:\mathrm{4x}\:-\:\mathrm{21}}\:\:+\:\mathrm{1}\:=\:\mathrm{0} \\ $$ Commented by Nicholas666 last updated on 14/May/25 $$…
Question Number 220473 by mehdee7396 last updated on 13/May/25 $${lim}_{{n}\rightarrow\infty} \left({tan}\left(\frac{\pi}{\mathrm{4}}+\frac{\mathrm{1}}{{n}}\right)\right)^{{n}} \:\:\:\:\:\overset{{t}=\frac{\mathrm{1}}{{n}}} {\rightarrow} \\ $$$$={lim}_{{t}\rightarrow\mathrm{0}} \left[{tan}\left(\frac{\pi}{\mathrm{4}}+{t}\right)\right]^{\frac{\mathrm{1}}{{t}}} \\ $$$$\Rightarrow{lim}_{{t}\rightarrow\mathrm{0}} \left[{tan}\left(\frac{\pi}{\mathrm{4}}+{t}\right)−\mathrm{1}\right]×\frac{\mathrm{1}}{{t}} \\ $$$$={lim}_{{t}\rightarrow\mathrm{0}} \left(\frac{\mathrm{1}+{tant}}{\mathrm{1}−{tant}}−\mathrm{1}\right)×\frac{\mathrm{1}}{{t}} \\ $$$$={lim}_{{t}\rightarrow\mathrm{0}} \left(\frac{\mathrm{2}{tant}}{\mathrm{1}−{tant}}\right)×\frac{\mathrm{1}}{{t}}=\mathrm{2}…
Question Number 220474 by Rojarani last updated on 13/May/25 Commented by Ghisom last updated on 13/May/25 $${a}_{{n}} =\mathrm{2}−\frac{\mathrm{1}}{{n}^{\mathrm{2}} +\sqrt{{n}^{\mathrm{4}} +\frac{\mathrm{1}}{\mathrm{4}}}}= \\ $$$$=\mathrm{4}{n}^{\mathrm{2}} +\mathrm{2}−\mathrm{2}\sqrt{\mathrm{4}{n}^{\mathrm{4}} +\mathrm{1}} \\…
Question Number 220468 by Rojarani last updated on 13/May/25 $$\:{a}+{b}+{c}=\mathrm{1},\:{a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} =\mathrm{1}\:\:\left({a},{b},{c}\:\in{R}\right) \\ $$$$\:{a}^{\mathrm{10}} +{b}^{\mathrm{10}} +{c}^{\mathrm{10}} =\mathrm{1},\:{a}^{\mathrm{4}} +{b}^{\mathrm{4}} +{c}^{\mathrm{4}} =? \\ $$ Answered by…
Question Number 220469 by Nicholas666 last updated on 13/May/25 $$ \\ $$$$\:\:\:\:\:\:\:\int_{\mathrm{0}} ^{\:\mathrm{1}} \:\:\frac{\mathrm{6}{x}\left(\mathrm{1}\:−\:{x}\right)}{\left({x}\:+\:\mathrm{1}\right)\left({x}^{\mathrm{2}} \:+\:\mathrm{1}\right)\:\mathrm{ln}\:\left({x}\:+\:\mathrm{1}\right)}\:{dx} \\ $$$$ \\ $$ Terms of Service Privacy Policy Contact:…
Question Number 220499 by Jubr last updated on 13/May/25 Commented by Jubr last updated on 13/May/25 $${Find}\:{x}\:{and}\:{y} \\ $$ Answered by Ghisom last updated on…
Question Number 220493 by leromain last updated on 13/May/25 $${Let}\:{ABC}\:{be}\:{a}\:{triangle}\:{such} \\ $$$${CosA}+{cosB}+{cosC}=\sqrt{\mathrm{2}} \\ $$$${SinA}+{sinB}+{sinC}=\mathrm{1}+\sqrt{\mathrm{2}} \\ $$$${Find}\:{A},{B},{C} \\ $$ Commented by mr W last updated on…
Question Number 220495 by Jubr last updated on 13/May/25 Commented by mr W last updated on 14/May/25 $${i}\:{guess}\:{even}\:{you}\:{also}\:{don}'{t}\:{know} \\ $$$${what}\:{the}\:{question}\:{means}\:{with}\:{all} \\ $$$${the}\:{arrows}.\:{when}\:{the}\:{question}\:{is} \\ $$$${unclear},\:{it}\:{can}\:{not}\:{be}\:{solved}.\:{so} \\…
Question Number 220486 by hardmath last updated on 13/May/25 $$\mathrm{z}\:\in\:\mathbb{C}\:\:\:\mathrm{and}\:\:\:\lambda\:>\:\mathrm{0} \\ $$$$\mathrm{Then}\:\mathrm{prove}\:\mathrm{that}: \\ $$$$\mid\mathrm{z}\:+\:\mathrm{2}\lambda\mid\:+\:\mid\mathrm{z}\:+\:\lambda\mid\:\geqslant\:\mid\mathrm{z}\:+\:\frac{\mathrm{3}\lambda\:−\:\lambda\sqrt{\mathrm{3}}\mathrm{i}}{\mathrm{2}}\mid \\ $$ Commented by MrGaster last updated on 14/May/25 The original problem is equivalent to: In a plane, the sum of the distances from any point to two vertices of an equilateral triangle is greater than the distance from that point to the third vertex. This can be easily proven. Then Ptolemy's theorem can be applied. Commented…
Question Number 220480 by SdC355 last updated on 13/May/25 $$\mathrm{Can}\:\mathrm{you}\:\mathrm{guys}\:\mathrm{teach}\:\mathrm{me}\:\mathrm{about} \\ $$$$\mathrm{Weber}\:\mathrm{function}\:\boldsymbol{\mathrm{E}}_{\nu} \left({z}\right)\:\mathrm{and}\:\mathrm{Anger}\:\mathrm{function}\:\boldsymbol{\mathrm{J}}_{\nu} \left({z}\right)?? \\ $$$$\: \\ $$$$\mathrm{Let}'\mathrm{s}\:\mathrm{Consider}\:{n}-\mathrm{dimensional}\:\mathrm{Euclidean}\:\mathrm{Space} \\ $$$$\mathrm{and}\:\mathrm{function}\:{f}\:,\:{f};\mathbb{R}^{{n}} \rightarrow\mathbb{R} \\ $$$$\mathrm{Helmholt}{z}\:\mathrm{Equation}\:\mathrm{defined}\:\mathrm{as} \\ $$$$\left(\bigtriangledown^{\mathrm{2}}…