Question

$\frac{1}{2+ \sqrt{5} } + \frac{1}{2- \sqrt{5} }$

Answer 1

 = (2 - R5 / (2 + R5)(2-R5)  ) + (2-R5 / (2 + R5)(2-R5)) = 
 ( 2+R5 + 2 - R5) / (2 + R5)(2-R5) = 4 / 2^2 - (R5)^2 =  4/(4 -5) = 4/-1 = -4  

Answer 2

   
[tex]\displaystyle\\ \frac{1}{2+ \sqrt{5} } + \frac{1}{2- \sqrt{5} }=\\\\ = \frac{1}{\sqrt{5}+2 } + \frac{1}{ -(\sqrt{5}-2) }=\\\\ =\frac{1}{\sqrt{5}+2 } - \frac{1}{\sqrt{5}-2 }=\\\\ ~~~~~~~~~~~~~~~~~~~~~\text{Numitorul comun este: }~ (\sqrt{5}+2 )(\sqrt{5}-2 )\\\\ = \frac{1\times (\sqrt{5}-2)}{(\sqrt{5}+2 )(\sqrt{5}-2)} - \frac{1\times (\sqrt{5}+2)}{(\sqrt{5}-2 )(\sqrt{5}+2)}=\\\\ = \frac{\sqrt{5}-2}{(\sqrt{5})^2-2^2 } - \frac{\sqrt{5}+2}{(\sqrt{5})^2-2^2}= [/tex]


[tex]\displaystyle\\ = \frac{\sqrt{5}-2}{5-4 } - \frac{\sqrt{5}+2}{5-4}=\\\\ = \frac{\sqrt{5}-2}{1} - \frac{\sqrt{5}+2}{1}=\\\\ =\sqrt{5}-2 -(\sqrt{5}+2)=\\\\ =\underline{\sqrt{5}}-2 -\underline{\sqrt{5}}-2= -2 -2 = \boxed{\bf -4} [/tex]