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 $$\mathrm{L}{et}\:{x}=\sqrt[{\mathrm{3}}]{\mathrm{2}+\sqrt{\mathrm{5}}}\:+\sqrt[{\mathrm{3}}]{\mathrm{2}−\sqrt{\mathrm{5}}}\: \\ $$$${We}\:{then}\:{have}\:{x}−\sqrt[{\mathrm{3}}]{\mathrm{2}+\sqrt{\mathrm{5}}}−\sqrt[{\mathrm{3}}]{\mathrm{2}−\sqrt{\mathrm{5}}}\:=\mathrm{0} \\…
Question Number 18808 by Bruce Lee last updated on 30/Jul/17 $$\sqrt[{\mathrm{3}}]{\mathrm{5}+\mathrm{2}\sqrt{\mathrm{5}}}\:+\:\sqrt[{\mathrm{3}}]{\mathrm{5}−\mathrm{2}\sqrt{\mathrm{5}}}\:=? \\ $$$$\boldsymbol{{help}}\:\boldsymbol{{me}} \\ $$ Answered by khamizan833@yahoo.com last updated on 30/Jul/17 Terms of Service…
Question Number 149876 by liberty last updated on 08/Aug/21 Answered by MJS_new last updated on 08/Aug/21 $${y}\geqslant\mathrm{0} \\ $$$${y}=\sqrt{{x}+\mathrm{2}\sqrt{{x}−\mathrm{1}}}+\sqrt{{x}−\mathrm{2}\sqrt{{x}−\mathrm{1}}} \\ $$$$\mathrm{squaring} \\ $$$${y}^{\mathrm{2}} =\left({x}+\mathrm{2}\sqrt{{x}−\mathrm{1}}\right)+\mathrm{2}\sqrt{\left({x}+\mathrm{2}\sqrt{{x}−\mathrm{1}}\right)\left({x}−\mathrm{2}\sqrt{{x}−\mathrm{1}}\right)}+\left({x}−\mathrm{2}\sqrt{{x}−\mathrm{1}}\right) \\…
Question Number 84341 by Rio Michael last updated on 11/Mar/20 $$\mathrm{Find}\:\mathrm{the}\:\mathrm{centre}\:\mathrm{of}\:\mathrm{symmetry}\:\mathrm{of}\:\mathrm{the} \\ $$$$\mathrm{curve}: \\ $$$$\:\:\:\:{y}\:=\:\frac{\mathrm{1}}{{x}\:+\:\mathrm{2}} \\ $$ Commented by mr W last updated on 11/Mar/20…
Question Number 18803 by daffa22 last updated on 30/Jul/17 Commented by daffa22 last updated on 30/Jul/17 $$\mathrm{help} \\ $$ Answered by behi.8.3.4.1.7@gmail.com last updated on…
Question Number 18801 by behi.8.3.4.1.7@gmail.com last updated on 30/Jul/17 Commented by behi.8.3.4.1.7@gmail.com last updated on 30/Jul/17 $${AB}=\mathrm{10},{AC}=\mathrm{12},{BC}=\mathrm{16} \\ $$$${connect}\:{foot}\:{of}\:{altitudes}\:{to}\:{make}\:{black} \\ $$$${triangles}\:{and}\:{do}\:{this}\:{in}\:{next}\:{triangles}, \\ $$$${as}\:{shown}. \\ $$$${find}\:{the}\:{area}\:{of}\:{last}\:{small}\:{black}\:{triangle}.…
Question Number 18800 by tawa tawa last updated on 29/Jul/17 $$\mathrm{Solve}:\:\:\mathrm{x}\:\mathrm{sin}\left(\mathrm{y}\right)\mathrm{dx}\:+\:\mathrm{x}^{\mathrm{2}} \:+\:\mathrm{y}\:\mathrm{cos}\left(\mathrm{y}\right)\mathrm{dy}\:=\:\mathrm{0},\:\:\mathrm{subject}\:\mathrm{to}\:\mathrm{y}\left(\mathrm{1}\right) \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 84334 by sahnaz last updated on 11/Mar/20 $$\int\frac{\mathrm{1}−\mathrm{u}}{−\mathrm{1}−\mathrm{2u}+\mathrm{u}^{\mathrm{2}} }\mathrm{du} \\ $$ Commented by niroj last updated on 11/Mar/20 $$\:\:\int\:\frac{−\left(\mathrm{u}−\mathrm{1}\right)}{−\left(\mathrm{1}+\mathrm{2u}−\mathrm{u}^{\mathrm{2}} \right)}\mathrm{du} \\ $$$$\:\:\:\int\:\frac{{u}−\mathrm{1}}{\mathrm{1}+\mathrm{2}{u}−{u}^{\mathrm{2}} }{du}…
Question Number 18799 by tawa tawa last updated on 29/Jul/17 Answered by Tinkutara last updated on 30/Jul/17 $$\mathrm{Diagonal}\:\mathrm{of}\:\mathrm{1}^{\mathrm{st}} \:\mathrm{square}\:=\:\frac{\mathrm{12}\sqrt{\mathrm{2}}}{\mathrm{2}} \\ $$$$\mathrm{Diagonal}\:\mathrm{of}\:\mathrm{2}^{\mathrm{nd}} \:\mathrm{square}\:=\:\frac{\mathrm{12}\sqrt{\mathrm{2}}}{\mathrm{4}} \\ $$$$\mathrm{Diagonal}\:\mathrm{of}\:\mathrm{3}^{\mathrm{rd}} \:\mathrm{square}\:=\:\frac{\mathrm{12}\sqrt{\mathrm{2}}}{\mathrm{8}}…
Question Number 149868 by mathdanisur last updated on 07/Aug/21 $$\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…