Question

Ajutati-ma va rog
a)Partea intreaga a numarului $\sqrt{3}$ este
b) Media geometrica a numerelor
a = $\sqrt{7-4 \sqrt{3} }$
b = $2 + \sqrt{3}$

Answer 1

a) 1
b) a=$\sqrt{7- 4\sqrt{3} }= \sqrt{ \frac{7+ \sqrt{49-48} }{2} } - \sqrt{ \frac{7- \sqrt{49-48} }{2} } = \sqrt{ \frac{8}{2} } - \sqrt{ \frac{6}{2} }=2- \sqrt{3}$
$\sqrt{a*b}= \sqrt{(2+ \sqrt{3})(2- \sqrt{3}) }= \sqrt{4-3} = \sqrt{1} =1$