Question

Fie a = 1+ [$\sqrt{2}$
b= $\sqrt{2}$ - 1  si mg - media geometrica a nr a si b. Aratati ca b < mg < a

Answer 1

Mg=√a.b
Mg=√(√2+1)(√2-1)=√2-1=√1=1
√2=1,41
1,41-1<1<1+1,41
Ceea ce este adevarat

Answer 2

a = √2 + 1
b = √2 - 1

mg = √(a * b) = √[(√2 + 1)(√2 - 1)] =√(2 - 1) = √1 = 1

b = 1,41 - 1 = 0,41  
a = 1,41 + 1 = 2,41
⇒ b < mg < a