Question Number 114809 by bemath last updated on 21/Sep/20 $$\begin{cases}{\mathrm{1}−\frac{\mathrm{12}}{{y}+\mathrm{3}{x}}=\frac{\mathrm{2}}{\:\sqrt{{x}}}}\\{\mathrm{1}+\frac{\mathrm{12}}{{y}+\mathrm{3}{x}}=\frac{\mathrm{6}}{\:\sqrt{{y}}}}\end{cases} \\ $$ Commented by bemath last updated on 21/Sep/20 $${set}\:\sqrt{{x}}\:=\:{u}\:\wedge\:\sqrt{{y}}\:=\:{v} \\ $$$$\Leftrightarrow\:\frac{\mathrm{12}}{{y}+\mathrm{3}{x}}\:=\:\frac{\mathrm{12}}{{y}+\mathrm{3}{x}}\:\Rightarrow\mathrm{1}−\frac{\mathrm{2}}{{u}}=\frac{\mathrm{6}}{{v}}−\mathrm{1} \\ $$$$\Rightarrow\:\mathrm{1}\:=\:\frac{\mathrm{1}}{{u}}+\frac{\mathrm{3}}{{v}};\:\mathrm{3}{u}+{v}\:=\:{uv} \\…
Question Number 49272 by rahul 19 last updated on 05/Dec/18 $${Z}\epsilon\mathbb{C}\:{satisfies}\:{the}\:{condition}\:\mid{Z}\mid\geqslant\mathrm{3}. \\ $$$${Then}\:{find}\:{the}\:{least}\:{value}\:{of}\:\mid{Z}+\frac{\mathrm{1}}{{Z}}\mid\:? \\ $$ Answered by tanmay.chaudhury50@gmail.com last updated on 05/Dec/18 Commented by tanmay.chaudhury50@gmail.com…
Question Number 114806 by Algoritm last updated on 21/Sep/20 Answered by MJS_new last updated on 21/Sep/20 $${a}−\mathrm{8}=\frac{\sqrt{\mathrm{24}}}{\:\sqrt{{a}}} \\ $$$$\mathrm{let}\:{a}={t}^{\mathrm{2}} \wedge{t}>\mathrm{0} \\ $$$${t}^{\mathrm{2}} −\mathrm{8}=\frac{\mathrm{2}\sqrt{\mathrm{6}}}{{t}} \\ $$$${t}^{\mathrm{3}}…
Question Number 114799 by Algoritm last updated on 21/Sep/20 Answered by MJS_new last updated on 21/Sep/20 $$\mathrm{you}\:\mathrm{can}\:\mathrm{only}\:\mathrm{try} \\ $$$$\mathrm{for}\:{x}\in\mathbb{N}\:\mathrm{the}\:\mathrm{only}\:\mathrm{solution}\:\mathrm{is}\:{x}=\mathrm{3} \\ $$ Terms of Service Privacy…
Question Number 114796 by Algoritm last updated on 21/Sep/20 Answered by MJS_new last updated on 21/Sep/20 $${y}=\frac{{x}}{\mathrm{2}+{y}}=\frac{{x}}{\mathrm{2}+\frac{{x}}{\mathrm{2}+{y}}}=\frac{{x}}{\mathrm{2}+\frac{{x}}{\mathrm{2}+\frac{{x}}{\mathrm{2}+{y}}}}=… \\ $$$$\Rightarrow\:{y}=−\mathrm{1}+\sqrt{{x}+\mathrm{1}} \\ $$$$\mathrm{but}\:\mathrm{the}\:\mathrm{last}\:\mathrm{term}\:\mathrm{1}+\sqrt{{x}+\mathrm{1}}=\mathrm{2}+{y} \\ $$$$\Rightarrow\:\mathrm{solution}\:\mathrm{is}\:−\mathrm{1}+\sqrt{{x}+\mathrm{1}} \\ $$…
Question Number 114795 by Algoritm last updated on 21/Sep/20 Answered by MJS_new last updated on 21/Sep/20 $$\mathrm{3}^{{x}} =\mathrm{27}−\mathrm{9}^{\mathrm{2}{y}} \\ $$$$\mathrm{2}^{{x}} =\frac{\mathrm{1}}{\mathrm{8}}−\mathrm{4}^{−{y}} \\ $$$$\Rightarrow \\ $$$$\left(\mathrm{1}\right)\:\mathrm{27}−\mathrm{9}^{\mathrm{2}{y}}…
Question Number 49256 by munnabhai455111@gmail.com last updated on 05/Dec/18 Answered by afachri last updated on 05/Dec/18 $$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{2}{x}^{\mathrm{2}} \:+\:\mathrm{3}{x}\:=\:\mathrm{10} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{x}^{\mathrm{2}} +\:\frac{\mathrm{3}}{\mathrm{2}}{x}\:=\:\mathrm{5}\:\:\:\:\:\:\:\:\:\:\Rightarrow\:\:\:\:\:\:\:\:{x}^{\mathrm{2}} \:+\:\mathrm{2}\left(\frac{\mathrm{3}}{\mathrm{4}}\right){x}\:=\:\mathrm{5} \\ $$$$\:\:\:\:\:{x}^{\mathrm{2}} \:+\:\mathrm{2}\left(\frac{\mathrm{3}}{\mathrm{4}}\right){x}\:+\:\left(\frac{\mathrm{3}}{\mathrm{4}}\right)^{\mathrm{2}}…
Question Number 49253 by cesar.marval.larez@gmail.com last updated on 04/Dec/18 Answered by afachri last updated on 05/Dec/18 $$\boldsymbol{\mathrm{no}}.\:\mathrm{2} \\ $$$$\boldsymbol{{f}}'\left({x}\right)\:=\:\:\underset{\boldsymbol{{h}}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\:\left[\frac{\:\:\:\:\:\left(\frac{\mathrm{3}\left({x}\:+\:{h}\right)\:+\:\mathrm{2}}{\mathrm{5}\left({x}\:+\:{h}\right)\:−\:\mathrm{2}}\:\:−\:\:\frac{\mathrm{3}{x}\:+\:\mathrm{2}}{\mathrm{5}{x}\:−\:\mathrm{2}\:}\right)\:\:\:\:}{{h}}\right] \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:=\:\underset{\boldsymbol{{h}}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\:\:\frac{\mathrm{1}}{{h}}\left(\:\frac{\:\mathrm{3}{x}\:+\:\mathrm{3}{h}\:+\:\mathrm{2}\:}{\mathrm{5}{x}\:+\:\mathrm{5}{h}\:−\:\mathrm{2}}\:\:−\:\:\frac{\:\mathrm{3}{x}\:+\:\mathrm{2}}{\mathrm{5}{x}\:−\:\mathrm{2}}\:\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:=\:\underset{\boldsymbol{{h}}\rightarrow\mathrm{0}}…
Question Number 49251 by cesar.marval.larez@gmail.com last updated on 04/Dec/18 Commented by afachri last updated on 05/Dec/18 $$\mathrm{unfortunately}\:\mathrm{i}\:\mathrm{am}\:\mathrm{not}\:\mathrm{learning}\:\mathrm{integral} \\ $$$$\mathrm{yet} \\ $$ Answered by tanmay.chaudhury50@gmail.com last…
Question Number 49249 by maxmathsup by imad last updated on 04/Dec/18 $${smplify}\:{A}_{{np}} =\sum_{{k}=\mathrm{0}} ^{{n}−\mathrm{1}} \:{cos}\left({pk}\right)\:\:{and}\:{B}_{{np}} \:=\sum_{{k}=\mathrm{0}} ^{{n}−\mathrm{1}} \:{sin}\left({pk}\right)\:{with}\:{p}\:{fromN} \\ $$ Terms of Service Privacy Policy…