Question

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Answer 1

Media aritmetica a numerelor a si b este:
$m_a=\frac{a+b}{2}=\frac{3\sqrt2 +2\sqrt3 +3\sqrt2 -2\sqrt3}{2}=\frac{6\sqrt2}{2}=3\sqrt2$

Media geometrica:
$m_g=\sqrt{a\times b}=\sqrt{(3\sqrt2 +2\sqrt3)(3\sqrt2 -2\sqrt3)}=\sqrt{(3\sqrt2)^2-(2\sqrt3)^2}=$
$=\sqrt{9\times 2-4\times 3}=\sqrt{18-12}=\sqrt6$

Raportul celor doua medii va fi:
$\frac{m_a}{m_g}=\frac{3\sqrt2}{\sqrt6}=\frac{3\sqrt2}{\sqrt{3} \times \sqrt{2}}=\frac{3}{\sqrt3}=\sqrt3$

Deci raspunsul corect este d) $\sqrt3$.

Answer 2

Răspuns:

Explicație pas cu pas:

a = 3√2 + 2√3

b = 3√2 - 2√3

media aritmetica = (a + b) : 2 = (3√2 + 2√3 + 3√2 - 2√3) : 2 = 6√2 : 2 = 3√2

a*b = (3√2 + 2√3)(3√2 - 2√3) = 18 - 12 = 6

media geometrica = √a*b = √6

media aritmetica/media geometrica = 3√2/√6 = 3√2*√6/6

= 3√12/6 = 3√4*3/6 = 3*2√3/6 = 6√3/6 = √3