Question Number 202466 by Rydel last updated on 27/Dec/23 $$\underset{{x}\rightarrow+\infty} {\mathrm{lim}}\frac{{x}\sqrt{\mathrm{ln}\:\left({x}^{\mathrm{2}} +\mathrm{1}\right)}}{\mathrm{1}+{e}^{{x}−\mathrm{3}} } \\ $$ Answered by Mathspace last updated on 27/Dec/23 $$={lim}_{{x}\rightarrow+\infty} {xe}^{−{x}+\mathrm{3}} \sqrt{{ln}\left(\mathrm{1}+{x}^{\mathrm{2}}…
Question Number 202393 by sonukgindia last updated on 26/Dec/23 Answered by Mathspace last updated on 26/Dec/23 $${I}=\int_{\mathrm{0}} ^{\infty} \frac{{lnx}}{{a}+{bx}^{\mathrm{2}} }{dx}\:\:\:\:\:\:\:\:\:\left({a}>\mathrm{0},{b}>\mathrm{0}\right) \\ $$$${I}=\frac{\mathrm{1}}{{a}}\int_{\mathrm{0}} ^{\infty} \frac{{lnx}}{\mathrm{1}+\frac{{b}}{{a}}{x}^{\mathrm{2}} }{dx}\:\:\left(\sqrt{\frac{{b}}{{a}}}{x}={t}\right)…
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Question Number 202447 by maths_plus last updated on 26/Dec/23 $$\mathrm{rationnalise}\:\mathrm{le}\:\mathrm{denominateur}\:\mathrm{de}\: \\ $$$$\mathrm{x}\:=\:\frac{\sqrt[{\mathrm{3}\:}]{\mathrm{2}}−\mathrm{1}}{\mathrm{1}−^{\mathrm{3}} \sqrt{\mathrm{2}}+^{\mathrm{3}} \sqrt{\mathrm{4}}} \\ $$ Answered by cortano12 last updated on 27/Dec/23 $$\:{a}^{\mathrm{3}} +{b}^{\mathrm{3}}…
Question Number 202295 by liuxinnan last updated on 24/Dec/23 $${psin}\theta{con}^{\mathrm{2}} \theta={a} \\ $$$${pcos}\theta{sin}^{\mathrm{2}} \theta={b} \\ $$$${p}\neq\mathrm{0}\:\:\:\:\theta\in\left(\mathrm{0}\:,\frac{\pi}{\mathrm{2}}\right) \\ $$$${prove}\:{p}=\frac{\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} \right)^{\frac{\mathrm{3}}{\mathrm{2}}} }{{ab}} \\ $$ Answered by…
Question Number 202287 by sonukgindia last updated on 24/Dec/23 Commented by a.lgnaoui last updated on 26/Dec/23 Answered by aleks041103 last updated on 24/Dec/23 $${I}\:{guess}\:{not}\:{enough}\:{information}. \\…
Question Number 202328 by MATHEMATICSAM last updated on 24/Dec/23 $$\mathrm{If}\:{n}\:\geqslant\:\mathrm{2}\:\mathrm{and}\:\mathrm{U}_{{n}} \:=\:\left(\mathrm{3}\:+\:\sqrt{\mathrm{5}}\right)^{{n}} \:+\:\left(\mathrm{3}\:−\:\sqrt{\mathrm{5}}\right)^{{n}} \\ $$$$\mathrm{then}\:\mathrm{prove}\:\mathrm{that}\:\mathrm{U}_{{n}\:+\:\mathrm{1}} \:=\:\mathrm{6U}_{{n}} \:−\:\mathrm{4U}_{{n}\:−\:\mathrm{1}} \:. \\ $$ Commented by aleks041103 last updated on…
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