Question Number 95693 by mathmax by abdo last updated on 27/May/20 $$\mathrm{solve}\:\mathrm{y}^{''} −\mathrm{2y}^{'} \:+\mathrm{1}\:=\left(\mathrm{x}−\mathrm{1}\right)\mathrm{shx} \\ $$ Answered by abdomathmax last updated on 29/May/20 $$\left(\mathrm{he}\right)\rightarrow\mathrm{r}^{\mathrm{2}} −\mathrm{2r}\:+\mathrm{1}\:=\mathrm{0}\:\Rightarrow\left(\mathrm{r}−\mathrm{1}\right)^{\mathrm{2}}…
Question Number 95695 by mathmax by abdo last updated on 27/May/20 $$\mathrm{solve}\:\left(\mathrm{x}+\mathrm{1}\right)\mathrm{y}^{'} −\mathrm{x}^{\mathrm{3}} \mathrm{y}\:=\:\mathrm{arctan}\left(\mathrm{2x}\right) \\ $$ Answered by mathmax by abdo last updated on 27/May/20…
Question Number 95662 by john santu last updated on 26/May/20 $$\mathrm{f}\left(\mathrm{x}+\mathrm{p}\right)\:+\:\mathrm{f}\left(\mathrm{x}−\mathrm{p}\right)\:=\:\mathrm{6x}−\mathrm{4} \\ $$$$\mathrm{f}\left(\mathrm{20}\right)\:=\:\mathrm{29p} \\ $$$$\frac{\mathrm{f}\left(\mathrm{p}\right)}{\mathrm{2p}}\:=\:? \\ $$ Commented by i jagooll last updated on 26/May/20…
Question Number 95634 by i jagooll last updated on 26/May/20 $$\mathrm{solve}\:\mid{x}+\frac{\mathrm{1}}{{x}}\mid\:>\:\mathrm{2}\: \\ $$ Answered by john santu last updated on 26/May/20 $$\mathrm{solution}:\: \\ $$$$\mathrm{this}\:\mathrm{is}\:\mathrm{equivalent}\:\mathrm{to}\:\mid\frac{{x}^{\mathrm{2}} +\mathrm{1}}{{x}}\mid\:>\:\mathrm{2}…
Question Number 30049 by abdo imad last updated on 15/Feb/18 $${find}\:\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{{n}\left({n}+\mathrm{1}\right)}{\mathrm{3}^{{n}} }\:. \\ $$ Commented by prof Abdo imad last updated on 16/Feb/18…
Question Number 161114 by cortano last updated on 12/Dec/21 $$\:\:{Let}\:{f}\left({x}\right)=\:\mathrm{sin}\:^{\mathrm{3}} \left(\mathrm{2}{x}\right)\:{for}\:−\frac{\pi}{\mathrm{4}}\leqslant{x}\leqslant\frac{\pi}{\mathrm{4}} \\ $$$$\:{then}\:{Df}^{−\mathrm{1}} \left(\frac{\mathrm{1}}{\mathrm{8}}\right)=\frac{{a}}{{b}\sqrt{{b}}}\:{so}\:\begin{cases}{{a}=?}\\{{b}=?}\end{cases} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 29987 by abdo imad last updated on 14/Feb/18 $${prove}\:{that}\:\:\sum_{{n}=\mathrm{1}_{{n}\neq{p}} } ^{\infty} \:\:\frac{\mathrm{1}}{{n}^{\mathrm{2}} \:−{p}^{\mathrm{2}} }\:\:=\:\:\frac{\mathrm{3}}{\mathrm{4}{p}^{\mathrm{2}} }\:. \\ $$ Terms of Service Privacy Policy Contact:…
Question Number 29986 by abdo imad last updated on 14/Feb/18 $${find}\:\sum_{{n}=\mathrm{0}} ^{\infty} \:\:\:\frac{{n}+\mathrm{1}}{\mathrm{4}^{{n}} }\:. \\ $$ Commented by abdo imad last updated on 14/Feb/18 $${let}\:{introduce}\:{for}\:\mid{x}\mid<\mathrm{1}\:\:\:{S}\left({x}\right)=\:\sum_{{n}=\mathrm{0}}…
Question Number 29985 by abdo imad last updated on 14/Feb/18 $${prove}\:{that}\:\:\:\:\sum_{{p}=\mathrm{1}} ^{\infty} \:\:\:\:\:\:\frac{{a}^{{p}} }{\mathrm{1}−{a}^{\mathrm{2}{p}} }\:=\:\sum_{{p}=\mathrm{1}} ^{\infty} \:\:\:\:\frac{{a}^{\mathrm{2}{p}−\mathrm{1}} }{\mathrm{1}−{a}^{\mathrm{2}{p}−\mathrm{1}} }\:. \\ $$ Terms of Service Privacy…
Question Number 29984 by abdo imad last updated on 14/Feb/18 $${prove}\:{that}\:\:\:\sum_{{n}=\mathrm{1}} ^{\infty} \:\:\:\frac{{H}_{{n}} }{{n}!}=={e}\:\sum_{{n}=\mathrm{1}} ^{\infty} \:\:\:\frac{\left(\mathrm{1}\right)^{\boldsymbol{{n}}−\mathrm{1}} }{\boldsymbol{{n}}\:\left(\boldsymbol{{n}}!\right)}\:. \\ $$ Terms of Service Privacy Policy Contact:…