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Please help

 Nov 30, 2016
 #1
avatar+245 
0

\(2( {-1+\sqrt5 \over4} )(\sqrt{(5+\sqrt5)\over8})\)

\(( {-1+\sqrt5 \over2} )(\sqrt{(5+\sqrt5)\over8})\)

\(( {-1+\sqrt5 \over2} )({\sqrt(5+\sqrt5)\over\sqrt8})\)

\(( {-1+\sqrt5 \over2} )({\sqrt(5+\sqrt5)\over2\sqrt2})\)

\({( {-1+\sqrt5 })({\sqrt(5+\sqrt5)})}\over4\sqrt2\)

\({ {-(\sqrt(5+\sqrt5)+\sqrt5*\sqrt(5+\sqrt5 })}\over4\sqrt2\)

\({ {-(\sqrt(5+\sqrt5)+\sqrt(5(5+\sqrt5) })}\over4\sqrt2\)

\({ {-(\sqrt(5+\sqrt5)+\sqrt(25+5\sqrt5) })}\over4\sqrt2\)

\({ {-(\sqrt5+\sqrt[4]{5}+5 +\sqrt5\sqrt[4]{5}) })}\over4\sqrt2\)

\({ {-(\sqrt5+\sqrt[4]{5}+5 +\sqrt[4]{5^3} })}\over4\sqrt2\)*This may be wrong*

 Dec 1, 2016
 #2
avatar+118677 
0

Let's see :)

 

\(2\left(\frac{-1+\sqrt5}{4}\right)\left(\sqrt{\frac{5+\sqrt5}{8}}\;\right)\\ =\left(\frac{-1+\sqrt5}{2}\right)\left(\frac {\sqrt{5+\sqrt5}}{\sqrt8}\;\right)\\ =\left(\frac{-1+\sqrt5}{2}\right)\left(\frac {\sqrt{5+\sqrt5}}{2\sqrt2}\;\right)\\ =\frac{(-1+\sqrt5) \sqrt{(5+\sqrt5) }}{4 \sqrt2 }\\ =\frac{\sqrt2(-1+\sqrt5) \sqrt{(5+\sqrt5) }}{8 }\\ =\frac{(\sqrt{10}-\sqrt2) \sqrt{(5+\sqrt5) }}{8 }\\\)

 

I haven't checked it. :/

 Dec 1, 2016
 #3
avatar+118677 
0

 

(\(\qquad (\sqrt{10}-\sqrt2)^2\\ \qquad =10+2-\sqrt{40}\\ \qquad=12-2\sqrt{10}\\ \\~\\ \frac{\sqrt{(12-2\sqrt{10})(5+\sqrt5)}}{8}\\ =\frac{\sqrt{60+12\sqrt5-10\sqrt{10}-2\sqrt{50}}}{8}\\ =\frac{\sqrt{60+12\sqrt5-10\sqrt{10}-10\sqrt{2}}}{8}\\\)

Melody  Dec 1, 2016
 #4
avatar+26393 
0

 Simplify c**p with radicals

 

 

 

\(\begin{array}{|rcll|} \hline && 2\cdot[~\frac14 (\sqrt{5}-1)~]\cdot \sqrt{\frac18\cdot (5+\sqrt{5}) } \\ &=& \frac24\cdot (\sqrt{5}-1)\cdot \sqrt{\frac18\cdot (5+\sqrt{5}) } \\ &=& \frac12\cdot (\sqrt{5}-1)\cdot \sqrt{\frac18\cdot (5+\sqrt{5}) } \\ &=& \frac12\cdot (\sqrt{5}-1)\cdot \frac{\sqrt{ 5+\sqrt{5} } } { \sqrt{8} } \\ &=& \frac{1}{2\cdot \sqrt{8} } \cdot (\sqrt{5}-1)\cdot \sqrt{ 5+\sqrt{5} } \\ &=& \frac{1}{2\cdot \sqrt{8} } \cdot \sqrt{ (\sqrt{5}-1)^2\cdot (5+\sqrt{5}) } \\ &=& \frac{1}{2\cdot \sqrt{8} } \cdot \sqrt{ (5-2\sqrt{5}+1)\cdot (5+\sqrt{5}) } \\ &=& \frac{1}{2\cdot \sqrt{8} } \cdot \sqrt{ (6-2\sqrt{5})\cdot (5+\sqrt{5}) } \\ &=& \frac{1}{2\cdot \sqrt{8} } \cdot \sqrt{ 30+6\sqrt{5}-10\sqrt{5}-2\cdot 5 } \\ &=& \frac{1}{2\cdot \sqrt{8} } \cdot \sqrt{ 30+6\sqrt{5}-10\sqrt{5}-10 } \\ &=& \frac{1}{2\cdot \sqrt{8} } \cdot \sqrt{ 20-4\sqrt{5} } \\ &=& \frac{1}{2\cdot \sqrt{8} } \cdot \sqrt{ 4(5-\sqrt{5}) } \\ &=& \frac{\sqrt{4}}{2\cdot \sqrt{8} } \cdot \sqrt{ 5-\sqrt{5} } \\ &=& \frac{2}{2\cdot \sqrt{8} } \cdot \sqrt{ 5-\sqrt{5} } \\ &=& \frac{1}{ \sqrt{8} } \cdot \sqrt{ 5-\sqrt{5} } \\ &=& \frac{1}{ \sqrt{4\cdot 2} } \cdot \sqrt{ 5-\sqrt{5} } \\ &=& \frac{1}{ \sqrt{4}\cdot \sqrt{2} } \cdot \sqrt{ 5-\sqrt{5} } \\ &=& \frac{1}{ 2\cdot \sqrt{2} } \cdot \sqrt{ 5-\sqrt{5} } \\ &=& \frac{\sqrt{2}} {\sqrt{2}}\cdot \frac{1}{ 2\cdot \sqrt{2} } \cdot \sqrt{ 5-\sqrt{5} } \\ &=& \frac{\sqrt{2}}{ 2\cdot 2 } \cdot \sqrt{ 5-\sqrt{5} } \\\\ &\mathbf{=}& \mathbf{ \frac{\sqrt{2}}{ 4 } \cdot \sqrt{ 5-\sqrt{5} } } \\ &=& 0.58778525229 \\ \hline \end{array}\)

 

 

laugh

 Dec 1, 2016

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