Question

Se considera numerele a=√98+√32-√50 si b=√162+√18-√72. Sa se afle media geometrica a numerelor a si b.

Answer 1

a=√98+√32-√50
a=7√2+4√2-5√2
a=11√2-5√2
a=6√2

b=√162+√18-√72
b=9√2+3√2-6√2
b=12√2-6√2
b=6√2

Mg=√a•b
Mg=√6√2•6√2
Mg=√36√4
Mg=√36•2
Mg=√72
Mg=6√2

Asta e tot,succes!

Answer 2

Media geometrica dintre 2 numere a, b este $\sqrt{a*b}$

Media geometrica dintre 3 numere a, b, c $\sqrt[2]{a*b*c}$

Revenind la exercitiu :
√98=7√2
√32=4√2
√50=5√2
Deci a=7√2+4√2-5√2=6√2

√162=9√2
√18=3√2
√72=6√2
Deci b=9√2+3√2-6√2=6√2

Media geometrica=$\sqrt{ 6\sqrt{2} * 6\sqrt{2} }$=6√2