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Category: Algebra

Suppose-a-b-c-are-nonzero-real-numbers-satisfying-ab-bc-ca-3-abc-a-b-c-3-Provd-that-a-b-c-must-be-terms-of-a-Geometric-Progession-

Question Number 125136 by Snail last updated on 08/Dec/20 $${Suppose}\:{a},{b},{c}\:{are}\:{nonzero}\:{real}\:{numbers} \\ $$$${satisfying}\:\left({ab}+{bc}+{ca}\right)^{\mathrm{3}} ={abc}\left({a}+{b}+{c}\right)^{\mathrm{3}} . \\ $$$${Provd}\:{that}\:{a},{b},{c}\:{must}\:{be}\:{terms}\:{of}\:{a}\:{Geometric} \\ $$$${Progession} \\ $$$$ \\ $$ Commented by Snail…

Let-a-b-c-complex-numbers-such-that-the-roots-of-the-equation-ax-2-bx-c-0-have-same-modulus-Prove-that-a-0-iff-b-0-

Question Number 125131 by Snail last updated on 03/Jun/21 $${Let}\:{a},{b},{c}\in\:{complex}\:{numbers}\:{such}\:{that}\:{the}\:{roots} \\ $$$${of}\:{the}\:{equation}\:{ax}^{\mathrm{2}} +{bx}+{c}=\mathrm{0}\:{have}\:{same}\:{modulus} \\ $$$${Prove}\:{that}\:{a}=\mathrm{0}\:{iff}\:{b}=\mathrm{0} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com

Prove-that-cos-pi-9-1-3-cos-2pi-9-1-3-cos-4pi-9-1-3-3-3-2-9-1-3-1-3-

Question Number 190659 by Shrinava last updated on 08/Apr/23 $$\mathrm{Prove}\:\mathrm{that}: \\ $$$$\sqrt[{\mathrm{3}}]{\mathrm{cos}\:\frac{\pi}{\mathrm{9}}}\:−\:\sqrt[{\mathrm{3}}]{\mathrm{cos}\:\frac{\mathrm{2}\pi}{\mathrm{9}}}\:−\:\sqrt[{\mathrm{3}}]{\mathrm{cos}\:\frac{\mathrm{4}\pi}{\mathrm{9}}} \\ $$$$=\:\sqrt[{\mathrm{3}}]{\mathrm{3}\:−\:\frac{\mathrm{3}}{\mathrm{2}}\:\sqrt[{\mathrm{3}}]{\mathrm{9}}}\: \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com

Question-125107

Question Number 125107 by peter frank last updated on 08/Dec/20 Commented by mr W last updated on 08/Dec/20 $${where}\:{did}\:{you}\:{get}\:{this}\:{question}? \\ $$$${it}\:{is}\:{wrong}.\:{for}\:{example}\:{m}=\mathrm{1},\:{n}=\mathrm{2}, \\ $$$${x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{2}=\mathrm{0}\:{has}\:{no}\:{real}\:{roots}! \\…

solve-in-R-1-x-x-2-

Question Number 190625 by mnjuly1970 last updated on 07/Apr/23 $$ \\ $$$$\:\:\:\:\:\:\:\:\:\mathrm{solve}\:\mathrm{in}\:\:\:\mathbb{R}\:\:\:: \\ $$$$\:\:\:\:\:\:\:\:\lfloor\:\frac{\mathrm{1}}{{x}}\:\rfloor\:\:+\:\lfloor\:{x}\:\rfloor\:=\:\mathrm{2}\:\:\:\:\: \\ $$$$ \\ $$ Answered by mr W last updated on…

1-4-1-4-1-8-

Question Number 59552 by hhghg last updated on 11/May/19 $$\frac{\mathrm{1}}{\mathrm{4}}+\left(\frac{\mathrm{1}}{\mathrm{4}}+\frac{\mathrm{1}}{\mathrm{8}}\right) \\ $$ Answered by ajfour last updated on 11/May/19 $$\frac{\mathrm{2}}{\mathrm{8}}+\frac{\mathrm{2}}{\mathrm{8}}+\frac{\mathrm{1}}{\mathrm{8}}=\frac{\mathrm{5}}{\mathrm{8}} \\ $$ Terms of Service…

9-5-4-5-3-

Question Number 59550 by hhghg last updated on 11/May/19 $$\mathrm{9}+\left(\mathrm{5}×\mathrm{4}+\mathrm{5}^{\mathrm{3}} \right) \\ $$ Answered by ajfour last updated on 11/May/19 $$\mathrm{20}+\mathrm{125}+\mathrm{9}=\mathrm{145}+\mathrm{9}=\mathrm{154} \\ $$ Terms of…