a) b=12[tex] \sqrt{3} = \sqrt{ 12^{2}*3 }= \sqrt{144*3} = \sqrt{432} [/tex]
[tex] \sqrt{b} [/tex] in cazul tau este [tex] \sqrt{432} [/tex].
Aplicam formula: [tex] \sqrt{ \frac{ 21 + \sqrt{441-432} }{2}} + \sqrt{ \frac{21 - \sqrt{441-432} }{2} }= [tex] \sqrt{ \frac{24}{2} }+ \sqrt{9} =2 \sqrt{3} +3[/tex]
b) Media geometrica= [tex] \sqrt{(2 \sqrt{3}-3)*( 2\sqrt{3} +3) } = \sqrt{12-9} = \sqrt{3} [/tex]. Am folosit formula (a-b)*(a+b)
