Question Number 194586 by Shrinava last updated on 10/Jul/23 $$\mathrm{abc}\:=\:\mathrm{e}^{\mathrm{3}} \:+\:\mathrm{d}^{\mathrm{3}} \:+\:\mathrm{f}^{\mathrm{3}} \\ $$$$\mathrm{edf}\:=\:\mathrm{a}^{\mathrm{3}} \:+\:\mathrm{b}^{\mathrm{3}} \:+\:\mathrm{c}^{\mathrm{3}} \\ $$$$\mathrm{find}:\:\mathrm{abc}\:\:\mathrm{and}\:\:\mathrm{edf}\: \\ $$ Commented by Shrinava last updated…
Question Number 194579 by MM42 last updated on 10/Jul/23 $${if}\:\:\:\:{u}_{{n}} =\frac{\mathrm{1}}{\:\sqrt{\mathrm{5}}}\left[\left(\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}\right)^{{n}} −\left(\frac{\mathrm{1}−\sqrt{\mathrm{5}}}{\mathrm{2}}\right)^{{n}} \right] \\ $$$$\:{then}\:\:\:{u}_{{n}+\mathrm{1}} ={u}_{{n}} +{u}_{{n}−\mathrm{1}} \:\:\:?\:\:\:\:\:;\:\:\:{n}=\mathrm{0},\mathrm{1},\mathrm{2},.. \\ $$ Answered by Frix last updated…
Question Number 194573 by Aduragbemi last updated on 10/Jul/23 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 194559 by MM42 last updated on 09/Jul/23 $${repeat}\:{question} \\ $$$${Shiw}\:{that}\:: \\ $$$$\underset{{i}=\mathrm{1}} {\overset{{n}} {\sum}}\:\left(\frac{\mathrm{1}}{\mathrm{2}{i}−\mathrm{1}}−\frac{\mathrm{1}}{\mathrm{2}{i}}\right)=\underset{{i}=\mathrm{1}} {\overset{{n}} {\sum}}\:\frac{\mathrm{1}}{{n}+{i}}\:\:? \\ $$ Answered by witcher3 last updated…
Question Number 194526 by mathlove last updated on 09/Jul/23 $$\frac{{f}\left({x}+\mathrm{1}\right)}{{f}\left({x}\right)}={x}^{\mathrm{2}\:\:\:\:\:\:\:\:} \:\:\:\:{f}\left({x}\right)=? \\ $$$$\frac{{f}\left(\mathrm{6}\right)}{{f}\left(\mathrm{3}\right)}=? \\ $$ Answered by JDamian last updated on 09/Jul/23 $$\frac{{f}\left(\mathrm{6}\right)}{{f}\left(\mathrm{3}\right)}=\frac{{f}\left(\mathrm{6}\right)}{{f}\left(\mathrm{3}\right)}×\frac{{f}\left(\mathrm{5}\right)}{{f}\left(\mathrm{5}\right)}×\frac{{f}\left(\mathrm{4}\right)}{{f}\left(\mathrm{4}\right)}= \\ $$$$=\frac{{f}\left(\mathrm{6}\right)}{{f}\left(\mathrm{5}\right)}×\frac{{f}\left(\mathrm{5}\right)}{{f}\left(\mathrm{4}\right)}×\frac{{f}\left(\mathrm{4}\right)}{{f}\left(\mathrm{3}\right)}=\mathrm{5}^{\mathrm{2}}…
Question Number 194522 by cortano12 last updated on 09/Jul/23 Answered by witcher3 last updated on 09/Jul/23 $$\mathrm{u}=\sqrt[{\mathrm{3}}]{\mathrm{x}−\mathrm{5}} \\ $$$$\mathrm{v}=\sqrt[{\mathrm{3}}]{\mathrm{7}−\mathrm{x}} \\ $$$$\mathrm{u}^{\mathrm{3}} +\mathrm{v}^{\mathrm{3}} =\mathrm{2p} \\ $$$$\frac{\mathrm{v}−\mathrm{u}}{\mathrm{u}+\mathrm{v}}=\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{v}^{\mathrm{3}}…
Question Number 194491 by horsebrand11 last updated on 08/Jul/23 $$\:\:\mathrm{x}=\sqrt{\mathrm{4}+\sqrt{\mathrm{5}\sqrt{\mathrm{3}}\:+\mathrm{5}\sqrt{\mathrm{48}−\mathrm{10}\sqrt{\mathrm{7}+\mathrm{4}\sqrt{\mathrm{3}}}}}}\: \\ $$$$\:\:\:\begin{array}{|c|}{\mathrm{2x}−\mathrm{1}=?}\\\hline\end{array} \\ $$ Answered by cortano12 last updated on 08/Jul/23 $$\:\:\mathrm{x}=\sqrt{\mathrm{4}+\sqrt{\mathrm{5}\sqrt{\mathrm{3}}+\mathrm{5}\sqrt{\mathrm{48}−\mathrm{10}\sqrt{\mathrm{7}+\mathrm{4}\sqrt{\mathrm{3}}}}}} \\ $$$$\:\:\mathrm{x}=\sqrt{\mathrm{4}+\sqrt{\mathrm{5}\sqrt{\mathrm{3}}+\mathrm{5}\sqrt{\mathrm{48}−\mathrm{10}\left(\mathrm{2}+\sqrt{\mathrm{3}}\right)}}} \\…
Question Number 194509 by pascal889 last updated on 08/Jul/23 Answered by horsebrand11 last updated on 09/Jul/23 $$\:\:\left.\begin{matrix}{\mathrm{u}=\mathrm{x}+\mathrm{2y}}\\{\mathrm{v}=\mathrm{2x}+\mathrm{y}}\end{matrix}\right\}\Rightarrow\mathrm{3}\left(\mathrm{x}+\mathrm{y}\right)=\mathrm{u}+\mathrm{v} \\ $$$$\:\Rightarrow\left.\begin{matrix}{\mathrm{u}+\mathrm{v}=\mathrm{uv}}\\{\frac{\mathrm{1}}{\mathrm{u}}+\frac{\mathrm{1}}{\mathrm{v}^{\mathrm{2}} }=\mathrm{3}}\end{matrix}\right\}\Rightarrow\begin{cases}{\mathrm{u}+\mathrm{v}^{\mathrm{2}} =\mathrm{3uv}^{\mathrm{2}} }\\{\mathrm{u}+\mathrm{v}=\mathrm{uv}}\end{cases} \\ $$$$\:\Rightarrow\begin{cases}{\mathrm{u}=\frac{\mathrm{v}^{\mathrm{2}} }{\mathrm{3v}^{\mathrm{2}}…
Question Number 194455 by Spillover last updated on 07/Jul/23 Answered by qaz last updated on 07/Jul/23 $$\frac{{x}^{\mathrm{4}} }{\mathrm{1}+{x}^{\mathrm{3}} }=\frac{{x}^{\mathrm{4}} }{{x}^{\mathrm{5}} +{x}^{\mathrm{3}} }=\frac{{x}}{\mathrm{1}+{x}^{\mathrm{2}} }\:\:\:\:\:\:\:,\frac{{x}^{\mathrm{3}} }{\mathrm{1}+{x}}=\frac{{x}^{\mathrm{3}} }{{x}^{\mathrm{5}}…
Question Number 194444 by Abdullahrussell last updated on 06/Jul/23 Answered by Frix last updated on 08/Jul/23 $$\mathrm{Let}\:{y}={px}\wedge{z}={qx} \\ $$$$\mathrm{It}'\mathrm{s}\:\mathrm{easy}\:\mathrm{to}\:\mathrm{get} \\ $$$${p}=\frac{{b}\left(\mathrm{4}−{b}\right)}{{ab}−\mathrm{2}\left({a}+{b}+{c}−\mathrm{4}\right)} \\ $$$${q}=\frac{{bc}−\mathrm{2}\left({a}+{b}+{c}−\mathrm{4}\right)}{{ab}−\mathrm{2}\left({a}+{b}+{c}−\mathrm{4}\right)} \\ $$$${ab}−\mathrm{2}\left({a}+{b}+{c}−\mathrm{4}\right)\neq\mathrm{0}…