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aratati ca media geometrica a numerelor a=2*3 si b=2*3 la puterea a 3a este cu 12 mai mica decat media lor aritmetica

Răspuns :

[tex]m_{g} = \sqrt{ab} = \sqrt{2 \cdot 3 \cdot 2 \cdot 3^{3} } = \sqrt{2^{2} \cdot 3^{4} } = \sqrt{2^{2} \cdot (3^{2})^{2} } = 2 \cdot 3^{2} = \bf 18\\[/tex]

[tex]m_{a} = \dfrac{2 \cdot 3 + 2 \cdot 3^{3} }{2} = \dfrac{6 + 2 \cdot 27}{2} = \dfrac{6 + 54}{2} = \dfrac{60}{2} = \bf 30\\[/tex]

[tex]m_{a} - m_{g} = 30 - 18 = \bf 12[/tex]

[tex]\it a=2\cdot3=6;\ \ \ b=2\cdot3^3=2\cdot27=54\\ \\ m_g=\sqrt{a\cdot b}=\sqrt{6\cdot54}=\sqrt{6\cdot6\cdot9}=\sqrt{36\cdot9}=6\cdot3=18\\ \\ m_a=\dfrac{a+b}{2}=\dfrac{6+54}{2}=\dfrac{60}{2}=30 =18+12=m_g+12 \Rightarrow m_a > m_g\ cu\ 12[/tex]