Question Number 180381 by Linton last updated on 11/Nov/22 $$\sqrt{\frac{\mathrm{8}^{\mathrm{12}} +\mathrm{4}^{\mathrm{13}} }{\mathrm{8}^{\mathrm{6}} +\mathrm{4}^{\mathrm{14}} }} \\ $$ Answered by floor(10²Eta[1]) last updated on 11/Nov/22 $$\sqrt{\frac{\mathrm{2}^{\mathrm{36}} +\mathrm{2}^{\mathrm{26}}…
Question Number 49279 by Tawa1 last updated on 05/Dec/18 Commented by MJS last updated on 05/Dec/18 $${x}−\mathrm{4}={u} \\ $$$$\mathrm{2}{x}−\mathrm{5}={v} \\ $$$${u}^{\mathrm{2017}} +{u}^{\mathrm{2015}} +{u}=−\left({v}^{\mathrm{2017}} +{v}^{\mathrm{2015}} +{v}\right)…
Question Number 180351 by MikeH last updated on 10/Nov/22 $$\mathrm{solve}\:\mathrm{the}\:\mathrm{differential}\:\mathrm{equation}\:\:{y}''\:+\mathrm{2}\:{y}'\:+\:{y}\:=\:\mathrm{0} \\ $$$$\mathrm{using}\:\mathrm{power}\:\mathrm{series}\:\mathrm{method}\:\mathrm{that}\:\mathrm{is},\:\mathrm{assume} \\ $$$${y}\:=\:\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}{c}_{{n}} {x}^{{n}} \:\mathrm{is}\:\mathrm{a}\:\mathrm{solution} \\ $$ Terms of Service Privacy Policy…
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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}}…