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Category: Algebra

nice-calculus-verify-that-A-2-5-1-3-1-5-is-a-rational-number-

Question Number 126109 by mnjuly1970 last updated on 17/Dec/20 $$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:….{nice}\:{calculus}… \\ $$$$\:\:{verify}\:\:{that}\:::\:\:\mathrm{A}=\frac{\sqrt[{\mathrm{3}}]{\mathrm{2}+\sqrt{\mathrm{5}}}}{\mathrm{1}+\sqrt{\mathrm{5}}}\:{is} \\ $$$$\:\:\:\:\:\:{a}\:\:{rational}\:\:{number}\:… \\ $$ Answered by som(math1967) last updated on 17/Dec/20 $$\frac{\sqrt[{\mathrm{3}}]{\mathrm{8}\left(\mathrm{2}+\sqrt{\left.\mathrm{5}\right)}\right.}}{\left(\mathrm{1}+\sqrt{\mathrm{5}}\right)×\mathrm{2}} \\…

1-a-b-c-d-2-1-a-1-b-1-c-1-d-abcd-4-

Question Number 126105 by Mathgreat last updated on 17/Dec/20 $$\mathrm{1}\leqslant{a};{b};{c};{d}\leqslant\mathrm{2} \\ $$$$\mid\left(\mathrm{1}−{a}\right)\left(\mathrm{1}−{b}\right)\left(\mathrm{1}−{c}\right)\left(\mathrm{1}−{d}\right)\mid\leqslant\frac{{abcd}}{\mathrm{4}} \\ $$ Commented by Mathgreat last updated on 17/Dec/20 $$\boldsymbol{{prove}} \\ $$ Answered…

Question-191624

Question Number 191624 by Shrinava last updated on 27/Apr/23 Answered by ARUNG_Brandon_MBU last updated on 28/Apr/23 $$\int_{\mathrm{1}} ^{{x}} \sqrt{\frac{{t}}{\mathrm{1}+{t}^{\mathrm{3}} }}{dt} \\ $$$$=\int_{\mathrm{1}} ^{{x}} \frac{{t}^{\frac{\mathrm{1}}{\mathrm{2}}} }{\:\sqrt{\mathrm{1}+\left({t}^{\frac{\mathrm{3}}{\mathrm{2}}}…

Question-191623

Question Number 191623 by Shrinava last updated on 27/Apr/23 Answered by mehdee42 last updated on 27/Apr/23 $${let}\::\:{f}\left({x}\right)={ax}+{b} \\ $$$${f}\left({f}\left({x}\right)\right)+{f}\left({x}\right)=−{x}\:\Rightarrow\left({a}^{\mathrm{2}} +{a}\right){x}+{ab}+\mathrm{2}{b}=−{x}\Rightarrow{a}=\frac{−\mathrm{1}+\sqrt{\mathrm{3}}{i}}{\mathrm{2}}={e}^{\frac{\mathrm{2}\pi}{\mathrm{3}}{i}} \:\:\&\:\:{b}=\mathrm{0}\Rightarrow{f}\left({x}\right)={e}^{\frac{\mathrm{2}\pi}{\mathrm{3}}{i}} {x} \\ $$$${f}\left({f}\left({f}\left(\frac{\mathrm{1}}{\mathrm{2}+{cosx}}\right)\right)\right)=\frac{\mathrm{1}}{\mathrm{2}+{cosx}} \\…

100-100-100-100-w-100-w-2-100-w-100-w-2-100-w-100-w-200-w-100-w-200-1-w-100-1-w-100-1-w-1-w-2

Question Number 191614 by vishal1234 last updated on 27/Apr/23 $$\frac{\alpha^{\mathrm{100}} +\beta^{\mathrm{100}} }{\alpha^{\mathrm{100}} −\beta^{\mathrm{100}} }\:=\: \\ $$$$\frac{\left(−{w}\right)^{\mathrm{100}} +\left(−{w}^{\mathrm{2}} \right)^{\mathrm{100}} }{\left(−{w}\right)^{\mathrm{100}} −\left(−{w}^{\mathrm{2}} \right)^{\mathrm{100}} } \\ $$$$=\:\frac{{w}^{\mathrm{100}} +{w}^{\mathrm{200}}…