Question

exercitiile 4 si 5 sau doar 4 va rog urgent

Answer 1

a) b=12$\sqrt{3} = \sqrt{ 12^{2}*3 }= \sqrt{144*3} = \sqrt{432}$

$\sqrt{b}$ in cazul tau este $\sqrt{432}$.

Aplicam formula: $\sqrt{ \frac{ 21 + \sqrt{441-432} }{2}} + \sqrt{ \frac{21 - \sqrt{441-432} }{2} }= [tex] \sqrt{ \frac{24}{2} }+ \sqrt{9} =2 \sqrt{3} +3$

b) Media geometrica= $\sqrt{(2 \sqrt{3}-3)*( 2\sqrt{3} +3) } = \sqrt{12-9} = \sqrt{3}$. Am folosit formula (a-b)*(a+b)