Question Number 42780 by maxmathsup by imad last updated on 02/Sep/18 $${find}\:{lim}_{{x}\rightarrow\mathrm{0}^{+} } \:\:\:\:\:\:\:\frac{\left[\left({x}+\mathrm{1}\right)^{\mathrm{2}} \right]\:−\left[\left(\mathrm{2}{x}+\mathrm{1}\right)^{\mathrm{2}} \right]}{{x}} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 42689 by prof Abdo imad last updated on 31/Aug/18 $${let}\:{f}\left({x}\right)\:=\:\frac{{x}}{{x}^{\mathrm{3}} −\mathrm{2}{x}\:\:+\mathrm{1}} \\ $$$$\left.\mathrm{1}\right)\:{find}\:{D}_{{f}} \\ $$$$\left.\mathrm{2}\right)\:{find}\:{f}^{\left({n}\right)} \left({x}\right)\:\:{then}\:\:{f}^{\left({n}\right)} \left(\mathrm{0}\right) \\ $$$$\left.\mathrm{3}\right)\:{developp}\:{f}\:{at}\:{integr}\:{serie}. \\ $$ Commented by…
Question Number 42688 by prof Abdo imad last updated on 31/Aug/18 $${let}\:{g}\left({x}\right)\:=\frac{{x}−\mathrm{1}}{{x}^{\mathrm{2}} +{x}\:+\mathrm{1}} \\ $$$$\left.\mathrm{1}\right)\:\:{find}\:{g}^{\left({n}\right)} \left({x}\right) \\ $$$$\left.\mathrm{2}\right){calculate}\:{g}^{\left({n}\right)} \left(\mathrm{0}\right) \\ $$$$\left.\mathrm{3}\right)\:{developp}\:{g}\:{at}\:\:{integr}\:{serie}. \\ $$ Commented by…
Question Number 108193 by bemath last updated on 15/Aug/20 $$\:\:\frac{\heartsuit\mathcal{B}{e}\mathcal{M}{ath}\heartsuit}{\blacklozenge} \\ $$$$\:{Given}\:\mathrm{2}{f}\left({x}\right)+\mathrm{3}{f}\left(\frac{\mathrm{1}}{{x}}\right)=\mathrm{2}{x}+\mathrm{3} \\ $$$${find}\:{f}\left(\mathrm{3}\right)\:?\: \\ $$ Commented by bemath last updated on 15/Aug/20 $${thank}\:{you}\:{both} \\…
Question Number 173678 by mathlove last updated on 16/Jul/22 $${if}\:{f}\left({x}\right)\:{is}\:\mathrm{2}^{{nd}} \:{digre}\:{function}\:\:\: \\ $$$${f}\left({x}−\mathrm{1}\right)+{f}\left({x}\right)+{f}\left({x}+\mathrm{1}\right)={x}^{\mathrm{2}} +\mathrm{1} \\ $$$${then}\:{faind}\:\:{f}\left(\mathrm{2}\right)=? \\ $$ Answered by Rasheed.Sindhi last updated on 16/Jul/22…
Question Number 42508 by maxmathsup by imad last updated on 26/Aug/18 $${find}\:{the}\:{value}\:{of}\: \\ $$$${A}\:={cos}\left(\frac{\pi}{\mathrm{5}}\right).{cos}\left(\frac{\mathrm{2}\pi}{\mathrm{5}}\right)\:{cos}\left(\frac{\mathrm{4}\pi}{\mathrm{5}}\right)\:{and}\:{B}\:={sin}\left(\frac{\pi}{\mathrm{5}}\right){sin}\left(\frac{\mathrm{2}\pi}{\mathrm{5}}\right){sin}\left(\frac{\mathrm{4}\pi}{\mathrm{5}}\right). \\ $$ Answered by math1967 last updated on 27/Aug/18 $${let}\frac{\pi}{\mathrm{5}}=\theta\:\:\therefore\mathrm{4}\theta=\pi−\theta \\…
Question Number 42493 by maxmathsup by imad last updated on 26/Aug/18 $$\:{calculate}\:{lim}_{{n}\rightarrow+\infty} \:\:\:\sum_{\mathrm{1}\leqslant{i}<{j}\leqslant{n}} \:\:\:\:\:\frac{\mathrm{1}}{{i}^{{x}} {j}^{{x}} }\:\:\:{with}\:\:{x}>\mathrm{1}\:\:{for}\:{that}\:{consider} \\ $$$$\xi\left({x}\right)\:=\sum_{{n}=\mathrm{1}} ^{\infty} \:\:\frac{\mathrm{1}}{{n}^{{x}} } \\ $$$$\left.\mathrm{2}\right)\:{calculate}\:{lim}_{{n}\rightarrow+\infty} \:\sum_{\mathrm{1}\leqslant{i}<{j}\leqslant{n}} \:\:\:\:\:\:\frac{\mathrm{1}}{\left({ij}\right)^{\mathrm{2}}…
Question Number 42492 by maxmathsup by imad last updated on 26/Aug/18 $${let}\:{x}>\mathrm{0}\:,{y}>\mathrm{0},{z}>\mathrm{0}\:\:\:{prove}\:{that}\:\:\frac{{x}^{\mathrm{2}} }{{yz}}\:+\frac{{y}^{\mathrm{2}} }{{xz}}\:+\frac{{z}^{\mathrm{2}} }{{xy}}\:\geqslant\mathrm{3}\:. \\ $$ Answered by tanmay.chaudhury50@gmail.com last updated on 26/Aug/18 $$\frac{{x}^{\mathrm{3}}…
Question Number 42482 by maxmathsup by imad last updated on 26/Aug/18 $${let}\:{f}\left({x}\right)={e}^{−\mathrm{2}{x}} \:{arctan}\left({x}\right) \\ $$$$\left.\mathrm{1}\right)\:{calculate}\:{f}^{\left({n}\right)} \left({x}\right) \\ $$$$\left.\mathrm{2}\right)\:{calculate}\:{f}^{\left({n}\right)} \left(\mathrm{0}\right) \\ $$$$\left.\mathrm{3}\right)\:{developp}\:{f}\:{at}\:{integr}\:{serie}\:. \\ $$ Commented by…
Question Number 42463 by abdo.msup.com last updated on 26/Aug/18 $${let}\:{y}\:\:=\sqrt{{x}+\sqrt{{x}+\sqrt{{x}+\mathrm{2}}}} \\ $$$${calculate}\:\:\frac{{dy}}{{dx}} \\ $$ Commented by maxmathsup by imad last updated on 26/Aug/18 $${we}\:{have}\:{y}^{\mathrm{2}} \:={x}+\sqrt{{x}+\sqrt{{x}+\mathrm{2}}\:}\:\:\Rightarrow\left({y}^{\mathrm{2}}…