Question Number 84384 by M±th+et£s last updated on 12/Mar/20 $$\left[{x}\right]^{{x}} =\mathrm{2}\sqrt{\mathrm{2}}\:\:,\:\forall{x}>\mathrm{0} \\ $$ Commented by mr W last updated on 12/Mar/20 $$\left[{x}\right]\leqslant{x} \\ $$$$\left[{x}\right]^{{x}} \leqslant{x}^{{x}}…
Question Number 149917 by mathdanisur last updated on 08/Aug/21 $$\underset{\:\mathrm{0}} {\overset{\:\mathrm{2}\boldsymbol{\pi}} {\int}}\frac{\mathrm{dt}}{\mathrm{4}\sqrt{\mathrm{2}}\:\mathrm{sin}\boldsymbol{\mathrm{t}}\:+\:\mathrm{6}}\:=\:? \\ $$ Answered by mathmax by abdo last updated on 08/Aug/21 $$\Psi=\int_{\mathrm{0}} ^{\mathrm{2}\pi}…
Question Number 84370 by Roland Mbunwe last updated on 12/Mar/20 $$\left.\mathrm{1}.\right)\:\mid{x}\mid\:+\mid{x}+\mathrm{2}\mid\:<\mathrm{5} \\ $$$$\left.\mathrm{2}.\right)\:\mid{x}\mid\:+\mid{x}+\mathrm{2}\mid\:+\:\mid\mathrm{2}−{x}\mid\:\leqslant\mathrm{8} \\ $$ Answered by john santu last updated on 12/Mar/20 $$\left.\mathrm{1}.\right)\:\left(\mathrm{i}\right)\:\mathrm{x}\geqslant\mathrm{0}\:\Rightarrow\:\mathrm{2x}\:<\:\mathrm{3}\:,\:\mathrm{x}<\frac{\mathrm{3}}{\mathrm{2}}\: \\…
Question Number 149889 by Samimsultani last updated on 08/Aug/21 Commented by bramlexs22 last updated on 08/Aug/21 $$\sqrt{\mathrm{a}+\mathrm{b}\sqrt{\mathrm{d}}}\:=\:\sqrt{\frac{\mathrm{a}+\mathrm{c}}{\mathrm{2}}}\:+\mathrm{sgn}\left(\mathrm{b}\right)\sqrt{\frac{\mathrm{a}−\mathrm{c}}{\mathrm{2}}} \\ $$$$\mathrm{where}\:\mathrm{c}=\sqrt{\mathrm{a}^{\mathrm{2}} −\mathrm{b}^{\mathrm{2}} \mathrm{d}}\: \\ $$ Commented by…
Question Number 149876 by liberty last updated on 08/Aug/21 Answered by MJS_new last updated on 08/Aug/21 $${y}\geqslant\mathrm{0} \\ $$$${y}=\sqrt{{x}+\mathrm{2}\sqrt{{x}−\mathrm{1}}}+\sqrt{{x}−\mathrm{2}\sqrt{{x}−\mathrm{1}}} \\ $$$$\mathrm{squaring} \\ $$$${y}^{\mathrm{2}} =\left({x}+\mathrm{2}\sqrt{{x}−\mathrm{1}}\right)+\mathrm{2}\sqrt{\left({x}+\mathrm{2}\sqrt{{x}−\mathrm{1}}\right)\left({x}−\mathrm{2}\sqrt{{x}−\mathrm{1}}\right)}+\left({x}−\mathrm{2}\sqrt{{x}−\mathrm{1}}\right) \\…
Question Number 149883 by liberty last updated on 08/Aug/21 $$\mathrm{Prove}\:\mathrm{that}\:\sqrt[{\mathrm{3}}]{\mathrm{2}+\sqrt{\mathrm{5}}}+\sqrt[{\mathrm{3}}]{\mathrm{2}−\sqrt{\mathrm{5}}}\:\mathrm{is} \\ $$$$\mathrm{a}\:\mathrm{rational}\:\mathrm{number} \\ $$ Answered by john_santu last updated on 08/Aug/21 $$\mathrm{L}{et}\:{x}=\sqrt[{\mathrm{3}}]{\mathrm{2}+\sqrt{\mathrm{5}}}\:+\sqrt[{\mathrm{3}}]{\mathrm{2}−\sqrt{\mathrm{5}}}\: \\ $$$${We}\:{then}\:{have}\:{x}−\sqrt[{\mathrm{3}}]{\mathrm{2}+\sqrt{\mathrm{5}}}−\sqrt[{\mathrm{3}}]{\mathrm{2}−\sqrt{\mathrm{5}}}\:=\mathrm{0} \\…
Question Number 149868 by mathdanisur last updated on 07/Aug/21 $$\mathrm{if}\:\:\mathrm{a};\mathrm{b}\:\:\mathrm{and}\:\:\mathrm{c}\:\:\mathrm{are}\:\mathrm{the}\:\mathrm{dimensions}\:\mathrm{of}\:\:\mathrm{a} \\ $$$$\mathrm{cuboid}\:\mathrm{with}\:\mathrm{the}\:\mathrm{diagonal}\:\boldsymbol{\mathrm{d}}\:\mathrm{then}\:\mathrm{prove} \\ $$$$\mathrm{d}\:\leqslant\:\sqrt{\frac{\mathrm{a}^{\mathrm{3}} }{\mathrm{b}}\:+\:\frac{\mathrm{b}^{\mathrm{3}} }{\mathrm{c}}\:+\:\frac{\mathrm{c}^{\mathrm{3}} }{\mathrm{a}}} \\ $$ Answered by dumitrel last updated on…
Question Number 18798 by chernoaguero@gmail.com last updated on 29/Jul/17 Answered by behi.8.3.4.1.7@gmail.com last updated on 30/Jul/17 $$\frac{{x}+{a}}{{x}+{b}}={t},\frac{{x}−{a}}{{x}−{b}}={s}\Rightarrow\frac{{x}^{\mathrm{2}} −{a}^{\mathrm{2}} }{{x}^{\mathrm{2}} −{b}^{\mathrm{2}} }={ts},\frac{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }{{ab}}={m} \\ $$$$\Rightarrow{t}^{\mathrm{2}}…
Question Number 149870 by mathdanisur last updated on 07/Aug/21 $$\mathrm{if}\:\:\:\mathrm{x};\mathrm{y};\mathrm{z};\mathrm{m};\mathrm{n}\in\mathbb{R}^{+} \:\:\mathrm{then}: \\ $$$$\underset{\boldsymbol{\mathrm{cyc}}} {\sum}\:\frac{\mathrm{b}^{−\mathrm{1}} }{\left(\mathrm{m}\sqrt{\mathrm{x}}\:+\:\mathrm{n}\sqrt{\mathrm{y}}\right)^{\mathrm{2}} }\:\geqslant\:\frac{\mathrm{3}}{\left(\mathrm{m}\:+\:\mathrm{n}\right)^{\mathrm{2}} } \\ $$ Commented by mathdanisur last updated on…
Question Number 84328 by sahnaz last updated on 11/Mar/20 $$\frac{\mathrm{3}}{\mathrm{7}}×\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}\left[\frac{\mathrm{u}−\mathrm{1}}{\mathrm{u}+\mathrm{1}}\right]−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}\left[\mathrm{u}^{\mathrm{2}} −\mathrm{1}\right] \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com