Question Number 105534 by I want to learn more last updated on 29/Jul/20 Commented by I want to learn more last updated on 29/Jul/20 $$\mathrm{the}\:\mathrm{bracket}\:\mathrm{is}\:\mathrm{a}\:\mathrm{floor}\:\mathrm{function}\:\mathrm{sir}.\:\:\mathrm{error}\:\mathrm{in}\:\mathrm{bracket}.…
Question Number 105536 by I want to learn more last updated on 29/Jul/20 Answered by adhigenz last updated on 29/Jul/20 $$\alpha+\beta\:=\:\mathrm{5},\:\alpha\beta\:=\:−\mathrm{2} \\ $$$${x}_{{n}} \:=\:\alpha^{{n}} −\beta^{{n}}…
Question Number 39995 by behi83417@gmail.com last updated on 14/Jul/18 Answered by tanmay.chaudhury50@gmail.com last updated on 15/Jul/18 $$\alpha+\beta={a}+{b}+\mathrm{1}\:\:\:\alpha\beta={a}+{b} \\ $$$$\alpha+\beta−\left({a}+{b}+\mathrm{1}\right)=\mathrm{0} \\ $$$$\alpha+\beta−\alpha\beta−\mathrm{1}=\mathrm{0} \\ $$$$\alpha\left(\mathrm{1}−\beta\right)−\left(\mathrm{1}−\beta\right)=\mathrm{0} \\ $$$$\left(\alpha−\mathrm{1}\right)\left(\mathrm{1}−\beta\right)=\mathrm{0}…
Question Number 171046 by Beginner last updated on 06/Jun/22 Answered by thfchristopher last updated on 07/Jun/22 $$\mathrm{Let}\:{u}={a}+{bx} \\ $$$${du}={bdx},\:{x}=\frac{{u}−{a}}{{b}} \\ $$$$\therefore\int\frac{{x}^{\mathrm{2}} }{\left({a}+{bx}\right)^{\mathrm{2}} }{dx} \\ $$$$=\frac{\mathrm{1}}{{b}^{\mathrm{3}}…
Question Number 39971 by math2018 last updated on 14/Jul/18 $${a}>\mathrm{0},{b}>\mathrm{0}, \\ $$$${What}\:{is}\:{the}\:{minimum}\:{value}\:{of} \\ $$$$\frac{{b}^{\mathrm{2}} +\mathrm{2}}{{a}+{b}}+\frac{{a}^{\mathrm{2}} }{{ab}+\mathrm{1}}\:\:\:? \\ $$ Answered by tanmay.chaudhury50@gmail.com last updated on 14/Jul/18…
Question Number 171043 by mathlove last updated on 06/Jun/22 $${A}\in{R} \\ $$$${A}=\frac{\sqrt{{x}−\mathrm{2}}+{x}+\mathrm{3}}{\:\sqrt{\mathrm{4}−\mathrm{2}{x}}+{x}−\mathrm{1}}\:\:\:\:\:\:\:\:\:\:{faind}\:{A}=? \\ $$ Answered by MJS_new last updated on 07/Jun/22 $$\sqrt{{x}−\mathrm{2}}\in\mathbb{R}\:\Rightarrow\:{x}\geqslant\mathrm{2} \\ $$$$\sqrt{\mathrm{4}−\mathrm{2}{x}}=\sqrt{\mathrm{2}}\sqrt{\mathrm{2}−{x}}\in\mathbb{R}\:\Rightarrow\:{x}\leqslant\mathrm{2} \\…
Question Number 171034 by mathlove last updated on 06/Jun/22 Commented by cortano1 last updated on 06/Jun/22 $$\left(\mathrm{12}\right)\sqrt{{a}+\sqrt{\mathrm{1}+\mathrm{2}}}\:=\:\mathrm{2} \\ $$$$\:\:\:\:\:\:\:\:\:\:{a}+\sqrt{\mathrm{3}}=\mathrm{4}\Rightarrow{a}=\mathrm{4}−\sqrt{\mathrm{3}} \\ $$$$\left(\mathrm{13}\right)\:\sqrt{{x}+\sqrt{\mathrm{2}+\mathrm{2}}}\:=\mathrm{2} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:{x}+\mathrm{2}=\mathrm{4}\Rightarrow{x}=\mathrm{2} \\ $$…
Question Number 105456 by 1549442205PVT last updated on 29/Jul/20 $$\mathrm{Prove}\:\mathrm{that}\:\mathrm{the}\:\mathrm{curve}\:\mathrm{y}=\mathrm{x}^{\mathrm{4}} +\mathrm{3x}^{\mathrm{2}} +\mathrm{2x} \\ $$$$\mathrm{does}\:\mathrm{not}\:\mathrm{meet}\:\mathrm{the}\:\mathrm{straight}\:\mathrm{line} \\ $$$$\mathrm{y}=\mathrm{2x}−\mathrm{1}\:\mathrm{and}\:\mathrm{find}\:\mathrm{the}\:\mathrm{distace}\:\mathrm{between} \\ $$$$\mathrm{their}\:\mathrm{nearest}\:\mathrm{points}.\left(\mathrm{Answer}\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{5}}}\right) \\ $$ Answered by ajfour last updated…
Question Number 105448 by bemath last updated on 29/Jul/20 $$\mathcal{G}{iven}\:\begin{cases}{\frac{\mathrm{4}{x}^{\mathrm{2}} }{\mathrm{4}{x}^{\mathrm{2}} +\mathrm{1}}\:=\:{y}}\\{\frac{\mathrm{4}{y}^{\mathrm{2}} }{\mathrm{4}{y}^{\mathrm{2}} +\mathrm{1}}\:=\:{z}}\\{\frac{\mathrm{4}{z}^{\mathrm{2}} }{\mathrm{4}{z}^{\mathrm{2}} +\mathrm{1}}\:=\:{x}}\end{cases}\:\:\:.\:{find} \\ $$$${x}+{y}+{z}\:?\: \\ $$ Commented by john santu last…
Question Number 170966 by Shrinava last updated on 04/Jun/22 $$\mathrm{If}\:\:\mathrm{a},\mathrm{b},\mathrm{c}>\mathrm{0}\:\:\mathrm{and}\:\:\mathrm{a}+\mathrm{b}+\mathrm{c}=\mathrm{3}\:\:\mathrm{then}: \\ $$$$\Sigma\:\frac{\mathrm{ab}}{\mathrm{a}^{\mathrm{2}} \:+\:\mathrm{2b}}\:\leqslant\:\mathrm{1} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com