Question Number 196621 by universe last updated on 28/Aug/23 $$\mathrm{find}\:\:\mathrm{f}\left(\mathrm{x}\right)\:\:\mathrm{if} \\ $$$$\:\mathrm{f}\left(\mathrm{x}+\sqrt{\mathrm{x}^{\mathrm{2}} +\mathrm{1}}\right)\:=\:\frac{\mathrm{x}}{\mathrm{x}+\mathrm{1}} \\ $$ Answered by MM42 last updated on 28/Aug/23 $${x}+\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}={u}\Rightarrow{x}=\frac{{u}^{\mathrm{2}} −\mathrm{1}}{\mathrm{2}{u}}…
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Question Number 196650 by CrispyXYZ last updated on 28/Aug/23 $$\mathrm{Prove}\:\mathrm{that}\:{x}^{\mathrm{3}} ={y}^{\mathrm{6}} +\mathrm{20}\:\mathrm{has}\:\mathrm{no}\:\mathrm{positive}\:\mathrm{integer}\:\mathrm{solution}. \\ $$ Answered by AST last updated on 28/Aug/23 $${x}^{\mathrm{3}} \equiv\mathrm{1},−\mathrm{1}\:{or}\:\mathrm{0}\left({mod}\:\mathrm{9}\right) \\ $$$${when}\:\left({y},\mathrm{9}\right)=\mathrm{1};{y}^{\mathrm{6}}…
Question Number 196619 by universe last updated on 28/Aug/23 Answered by cortano12 last updated on 28/Aug/23 $$\:\:\mathrm{f}\left(\mathrm{x}+\sqrt{\mathrm{x}^{\mathrm{2}} +\mathrm{1}}\:\right)\:=\:−\frac{\mathrm{1}}{\mathrm{x}+\sqrt{\mathrm{x}^{\mathrm{2}} +\mathrm{1}}} \\ $$$$\:\:\mathrm{f}\left(\mathrm{x}\right)=\:−\frac{\mathrm{1}}{\mathrm{x}} \\ $$$$\:\:\mathrm{f}\left(\mathrm{2}\right)+\mathrm{f}\left(\mathrm{3}\right)+\mathrm{f}\left(\mathrm{6}\right)\:=\:−\left(\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{6}}\right)\:=\:\begin{array}{|c|}{−\mathrm{1}}\\\hline\end{array} \\ $$$$…
Question Number 196612 by Calculusboy last updated on 28/Aug/23 Commented by Frix last updated on 28/Aug/23 $$\mathrm{Question}\:\mathrm{196165} \\ $$ Terms of Service Privacy Policy Contact:…
Question Number 196614 by mr W last updated on 28/Aug/23 $${if}\:{S}_{{n}} =\frac{\mathrm{1}}{\mathrm{1}+\mathrm{5}{n}}+\frac{\mathrm{1}}{\mathrm{2}+\mathrm{5}{n}}+\frac{\mathrm{1}}{\mathrm{3}+\mathrm{5}{n}}+…+\frac{\mathrm{1}}{\mathrm{6}{n}}, \\ $$$${find}\:\underset{{n}\rightarrow\infty} {\mathrm{lim}}{S}_{{n}} =? \\ $$ Answered by universe last updated on 28/Aug/23…
Question Number 196540 by Matica last updated on 27/Aug/23 $$\:\:{using}\:{definition}\:{of}\:{limit},\:{prove}\: \\ $$$$\:\:\:\underset{{x}\rightarrow+\infty} {\mathrm{lim}}\frac{{x}}{{x}+\mathrm{1}}\:=\:\mathrm{1} \\ $$ Answered by ERLY last updated on 27/Aug/23 $${tu}\:{appliques}\:{le}\:{monome}\:{de}\:{plus}\:{haut}\:{degre}\:{tu}\:{aura}\:{lim}\:{x}/{x}\:{ce}\:{qui}\:{donne}\:\mathrm{1}\:\:\:\:\:\:{Erly}\:{rolvinst} \\ $$…
Question Number 196539 by qaz last updated on 27/Aug/23 $$\underset{{i}=\mathrm{1}} {\overset{\mathrm{6}} {\sum}}{f}\left({i}\right)=\underset{{i}=\mathrm{1}} {\overset{\mathrm{6}} {\sum}}{f}\left(\mathrm{6}+\mathrm{1}−{i}\right) \\ $$$$\underset{{i}=\mathrm{1}} {\overset{\mathrm{6}} {\sum}}\underset{{j}=\mathrm{1}} {\overset{{i}} {\sum}}{f}\left({i},{j}\right)=\underset{{i}=\mathrm{1}} {\overset{\mathrm{6}} {\sum}}\underset{{j}=\mathrm{1}} {\overset{{i}} {\sum}}{f}\left(\mathrm{6}+\mathrm{1}−{j},\mathrm{6}+\mathrm{1}−{i}\right) \\…
Question Number 196596 by MATHEMATICSAM last updated on 27/Aug/23 $$\mathrm{If}\:{f}\left({x}\right)\:=\:\mathrm{ln}\left(\frac{\mathrm{1}\:+\:{x}}{\mathrm{1}\:−\:{x}}\right)\:\mathrm{then}\:\mathrm{prove}\:\mathrm{that} \\ $$$${f}\left(\frac{\mathrm{2}{x}}{\mathrm{1}\:+\:{x}^{\mathrm{2}} }\right)\:=\:\mathrm{2}{f}\left({x}\right). \\ $$ Commented by mokys last updated on 27/Aug/23 $${f}\:\left(\frac{\mathrm{2}{x}}{\mathrm{1}+{x}^{\mathrm{2}} }\right)=\:{ln}\left(\frac{\mathrm{1}+\frac{\mathrm{2}{x}}{\mathrm{1}+{x}^{\mathrm{2}} }}{\mathrm{1}−\frac{\mathrm{2}{x}}{\mathrm{1}+{x}^{\mathrm{2}}…
Question Number 196534 by cortano12 last updated on 27/Aug/23 $$\:\:\:\:\:\:\:\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\mathrm{x}\:\sqrt{\mathrm{1}−\mathrm{cos}\:\left(\frac{\pi}{\mathrm{x}}\right)}\:=? \\ $$ Answered by Mathspace last updated on 27/Aug/23 $$\mathrm{1}−{cosu}\sim\frac{{u}^{\mathrm{2}} }{\mathrm{2}}\:\left({u}\rightarrow{o}\right)\:\Rightarrow \\ $$$$\mathrm{1}−{cos}\left(\frac{\pi}{{x}}\right)\sim\frac{\pi^{\mathrm{2}} }{\mathrm{2}{x}^{\mathrm{2}}…