Răspunsuri

2014-06-17T19:33:46+03:00
Prin formula radicalilor dubli
 \sqrt{3+ \sqrt{5} }  
3=a
5=b
c= \sqrt{a^{2}-b }  \sqrt{9-5} =√4=2
 \sqrt{3+ \sqrt{5} }  =  \sqrt{ \frac{a+c}{2} } +   \sqrt{ \frac{a-c}{2} }  = \sqrt{ \frac{3+2}{2} } + \sqrt{  \frac{3-2}{2} } =  \sqrt{ \frac{5}{2} } +   \sqrt{ \frac{1}{2} } , iar   \sqrt{3- \sqrt{5} }  =  \sqrt{ \frac{5}{2} } -  \sqrt{ \frac{1}{2} } x= \sqrt{ \frac{5}{2} } + \sqrt{ \frac{1}{2} } + \sqrt{ \frac{5}{2} } - \sqrt{ \frac{1}{2} }= 2 \sqrt{ \frac{5}{2} }


 \sqrt{10} - 2 \sqrt{ \frac{5}{2} } -1 =\sqrt{10} -2 \sqrt{ \frac{25}{100} } -1 =\sqrt{10} - 2 *\frac{5}{10} -1 = \sqrt{10} - \frac{10}{10} -1 =   \sqrt{10} - 1 -1 =  \sqrt{10} -2