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 $$\mathrm{Let}\:{y}={px}\wedge{z}={qx} \\ $$$$\mathrm{It}'\mathrm{s}\:\mathrm{easy}\:\mathrm{to}\:\mathrm{get} \\ $$$${p}=\frac{{b}\left(\mathrm{4}−{b}\right)}{{ab}−\mathrm{2}\left({a}+{b}+{c}−\mathrm{4}\right)} \\ $$$${q}=\frac{{bc}−\mathrm{2}\left({a}+{b}+{c}−\mathrm{4}\right)}{{ab}−\mathrm{2}\left({a}+{b}+{c}−\mathrm{4}\right)} \\ $$$${ab}−\mathrm{2}\left({a}+{b}+{c}−\mathrm{4}\right)\neq\mathrm{0}…
Question Number 194422 by York12 last updated on 06/Jul/23 $${What}\:{books}\:{use}\:{for}\:{studying}\:{inequalities} \\ $$$${for}\:{beginners}\: \\ $$ Answered by ajfour last updated on 06/Jul/23 $${Elementary}\:{Mathematics} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:-\:{G}.\:{Dorofeev} \\…
Question Number 194389 by mathlove last updated on 05/Jul/23 $$\sqrt{\sqrt{\mathrm{49}+\mathrm{20}\sqrt{\mathrm{6}}}}=? \\ $$ Answered by som(math1967) last updated on 05/Jul/23 $$\sqrt{\sqrt{\mathrm{5}^{\mathrm{2}} +\left(\mathrm{2}\sqrt{\mathrm{6}}\right)^{\mathrm{2}} +\mathrm{2}.\mathrm{5}.\mathrm{2}\sqrt{\mathrm{6}}}} \\ $$$$\sqrt{\sqrt{\left(\mathrm{5}+\mathrm{2}\sqrt{\mathrm{6}}\right)^{\mathrm{2}} }}…
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Question Number 194373 by MM42 last updated on 05/Jul/23 $${Show}\:\:\: \\ $$$$\underset{{i}=\mathrm{1}} {\overset{{n}} {\sum}}\left(\frac{\mathrm{1}}{\mathrm{2}{i}−\mathrm{1}}−\:\frac{\mathrm{1}}{\mathrm{2}{i}}\right)=\underset{{i}=\mathrm{1}} {\overset{{n}} {\sum}}\:\frac{\mathrm{1}}{{n}+{i}}\: \\ $$$$ \\ $$ Terms of Service Privacy Policy…
Question Number 194326 by horsebrand11 last updated on 03/Jul/23 $$\:\:\sqrt{\frac{\sqrt{\mathrm{x}^{\mathrm{2}} +\mathrm{66}^{\mathrm{2}} +\mathrm{x}}}{\mathrm{x}}\:}\:−\sqrt{\mathrm{x}\sqrt{\mathrm{x}^{\mathrm{2}} +\mathrm{66}^{\mathrm{2}} }−\mathrm{x}^{\mathrm{2}} }\:=\:\mathrm{5}\: \\ $$ Answered by cortano12 last updated on 03/Jul/23 $$\:\:\:\cancel{\lesseqgtr}…