Category: Algebra

Question Number 90473 by I want to learn more last updated on 23/Apr/20 Commented by MJS last updated on 24/Apr/20 you want to learn more. I want to know if you have got any idea what a "proper ideal" might be. Commented by I want…

Question Number 24908 by Tinkutara last updated on 28/Nov/17 $$\mathrm{If}\:{a},\:{b},\:{c}\:\mathrm{are}\:\mathrm{the}\:\mathrm{sides}\:\mathrm{of}\:\mathrm{a}\:\mathrm{triangle}\:\mathrm{prove} \\ $$$$\mathrm{the}\:\mathrm{following}\:\mathrm{inequality}: \\ $$$$\frac{{a}}{{c}\:+\:{a}\:−\:{b}}\:+\:\frac{{b}}{{a}\:+\:{b}\:−\:{c}}\:+\:\frac{{c}}{{b}\:+\:{c}\:−\:{a}}\:\geqslant\:\mathrm{3}. \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com

Question Number 155979 by mnjuly1970 last updated on 06/Oct/21 $$ \\ $$$${prove}\:{that}: \\ $$$$\int_{\mathrm{0}} ^{\:\mathrm{1}} \int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{\sqrt{{x}}\:+\sqrt{{y}}}{\left(\mathrm{1}−{xy}\right)\sqrt[{\mathrm{4}}]{{xy}}}\:{dxdy}=\mathrm{4}\:\left(\mathrm{4}\:−\pi\right) \\ $$$$ \\ $$ Terms of Service…

Question Number 155969 by mathlove last updated on 06/Oct/21 Answered by MJS_new last updated on 06/Oct/21 $${a}=\sqrt[{\mathrm{2014}}]{\mathrm{2}^{\mathrm{0}} }+\sqrt[{\mathrm{2014}}]{\mathrm{2}^{\mathrm{1}} }+…+\sqrt[{\mathrm{2014}}]{\mathrm{2}^{\mathrm{2013}} }= \\ $$$$=\underset{{n}=\mathrm{0}} {\overset{\mathrm{2013}} {\sum}}\mathrm{2}^{\frac{{n}}{\mathrm{2014}}} \\…

Question Number 155951 by mnjuly1970 last updated on 06/Oct/21 $$ \\ $$$$\:\:\:\:\:\:\:{solve}\:\:: \\ $$$$ \\ $$$$\:\:\:\:\:\lfloor\:\frac{\mathrm{1}}{{x}}\:\rfloor\:+\:\lfloor\:\frac{\mathrm{3}}{{x}}\:\rfloor=\:\mathrm{4}\: \\ $$$$ \\ $$ Answered by mr W last…