Question Number 66755 by John Kaloki Musau last updated on 19/Aug/19 $${solve}\:{the}\:{simultaneous}\:{equations}: \\ $$$$\mathrm{3}{x}-{y}=\mathrm{9} \\ $$$${x}^{\mathrm{2}} -{xy}=\mathrm{4} \\ $$ Answered by $@ty@m123 last updated on…
Question Number 66727 by mathmax by abdo last updated on 19/Aug/19 $${let}\:\:{U}_{{n}} =\sum_{{k}=\mathrm{0}} ^{{n}} \:\frac{\mathrm{1}}{\mathrm{3}{k}+\mathrm{1}}\:\:\:{and}\:{H}_{{n}} =\sum_{{k}=\mathrm{1}} ^{{n}} \:\frac{\mathrm{1}}{{k}} \\ $$$${calculate}\:{U}_{{n}} \:{interms}\:{of}\:{H}_{{n}} \\ $$ Terms of…
Question Number 1180 by 934111 last updated on 11/Jul/15 $${f}\left({x}^{\mathrm{2}} \right)−{f}\left({x}\right)=\mathrm{1} \\ $$ Commented by 123456 last updated on 11/Jul/15 $$\left\{\mathrm{0},\mathrm{1}\right\}\notin\mathrm{D}\left({f}\right) \\ $$ Answered by…
Question Number 66696 by mathmax by abdo last updated on 18/Aug/19 $${calculate}\:{lim}_{{x}\rightarrow\mathrm{0}} \:\:\:\:\:\frac{{arctan}\left(\mathrm{1}+{x}^{\mathrm{3}} \right)−\frac{\pi}{\mathrm{4}}}{{xsin}\left({x}^{\mathrm{2}} \right)} \\ $$ Commented by kaivan.ahmadi last updated on 18/Aug/19 $${lim}_{{x}\rightarrow\mathrm{0}}…
Question Number 66680 by mathmax by abdo last updated on 18/Aug/19 $${calculate}\:\:\sum_{{n}=\mathrm{1}} ^{\infty} \:\:\:\:\frac{\mathrm{2}^{{n}} }{\mathrm{3}^{{n}} \left(\mathrm{2}{n}^{\mathrm{3}} \:+{n}^{\mathrm{2}} −\mathrm{5}{n}\:+\mathrm{2}\right)} \\ $$ Commented by Mohamed Amine Bouguezzoul…
Question Number 1120 by 123456 last updated on 17/Jun/15 $$\mathrm{given}\:{f}:\mathbb{R}\rightarrow\mathbb{R} \\ $$$${f}\left({x}\right)={a}_{{n}} {x}^{{n}} +\centerdot\centerdot\centerdot+{a}_{\mathrm{0}} \: \\ $$$$\mathrm{with}\:{a}_{{n}} \neq\mathrm{0},\:{a}_{{i}} \in\mathbb{R},{i}\in\left\{\mathrm{0},\mathrm{1},…,{n}\right\},{n}\in\mathbb{N}^{\ast} \\ $$$$\mathrm{supose}\:\mathrm{that}\:{x}_{\mathrm{1}} ,…,{x}_{{n}} \\ $$$$\mathrm{are}\:\mathrm{roots}\:\mathrm{the}\:{n}\:\mathrm{roots}\:\mathrm{of}\:{f}\left({x}\right) \\…
Question Number 1117 by 123456 last updated on 17/Jun/15 $${a}_{\mathrm{0}} =\mathrm{1} \\ $$$${a}_{{n}} =\frac{\sqrt{{a}_{{n}−\mathrm{1}} }}{{a}_{{n}−\mathrm{1}} }+\mathrm{1} \\ $$$$\underset{{n}\rightarrow+\infty} {\mathrm{lim}}\:{a}_{{n}} =? \\ $$ Commented by 123456…
Question Number 1088 by 112358 last updated on 10/Jun/15 $${Solve}\:{the}\:{following}\:{integral} \\ $$$${equation}\:{for}\:{f}\left({x}\right): \\ $$$$\int_{\mathrm{0}} ^{\:{x}} {f}\left({t}\right){dt}=\mathrm{3}{f}\left({x}\right)+{k} \\ $$$${where}\:{k}\:{is}\:{a}\:{constant}.\: \\ $$ Answered by prakash jain last…
Question Number 132135 by benjo_mathlover last updated on 11/Feb/21 Answered by Olaf last updated on 11/Feb/21 $${f}\left({x}\right)+{f}\left(\frac{{x}−\mathrm{1}}{{x}}\right)\:=\:\mathrm{1}+{x}\:\:\:\:\:\left(\mathrm{1}\right) \\ $$$$\mathrm{Let}\:{x}\:=\:\frac{{u}−\mathrm{1}}{{u}} \\ $$$$\left(\mathrm{1}\right)\::\:{f}\left(\frac{{u}−\mathrm{1}}{{u}}\right)+{f}\left(\frac{\frac{{u}−\mathrm{1}}{{u}}−\mathrm{1}}{\frac{{u}−\mathrm{1}}{{u}}}\right)\:=\:\mathrm{1}+\frac{{u}−\mathrm{1}}{{u}} \\ $$$${f}\left(\frac{{u}−\mathrm{1}}{{u}}\right)+{f}\left(\frac{\mathrm{1}}{\mathrm{1}−{u}}\right)\:=\:\mathrm{1}+\frac{{u}−\mathrm{1}}{{u}}\:\:\:\:\:\left(\mathrm{2}\right) \\ $$$$\left(\mathrm{1}\right)−\left(\mathrm{2}\right)\::…
Question Number 950 by 123456 last updated on 06/May/15 $${f}:\mathbb{R}\rightarrow\mathbb{R},\mathrm{continuous}\:\mathrm{such}\:\mathrm{that}\:\forall{x}\in\mathbb{R},{f}\left({x}\right){f}\left[{f}\left({x}\right)\right]=\mathrm{1},{f}\left(\mathrm{2004}\right)=\mathrm{2003},{f}\left(\mathrm{1999}\right)=? \\ $$ Commented by 123456 last updated on 06/May/15 $${f}\left(\mathrm{2004}\right){f}\left[{f}\left(\mathrm{2004}\right)\right]=\mathrm{2003}{f}\left(\mathrm{2003}\right)=\mathrm{1}\Leftrightarrow{f}\left(\mathrm{2003}\right)=\frac{\mathrm{1}}{\mathrm{2003}} \\ $$$${f}\left(\mathrm{2003}\right){f}\left[{f}\left(\mathrm{2003}\right)\right]=\frac{\mathrm{1}}{\mathrm{2003}}{f}\left(\frac{\mathrm{1}}{\mathrm{2003}}\right)=\mathrm{1}\Leftrightarrow{f}\left(\frac{\mathrm{1}}{\mathrm{2003}}\right)=\mathrm{2003}={f}\left(\mathrm{2004}\right) \\ $$$${f}\left(\frac{\mathrm{1}}{\mathrm{2003}}\right){f}\left[{f}\left(\frac{\mathrm{1}}{\mathrm{2003}}\right)\right]=\mathrm{2003}{f}\left(\mathrm{2003}\right)=\mathrm{1} \\…