Question

Clasa a 7-a dau coroana!
Calculeaza media geometrica a numerelor reale x si y in urmatoarele cazuri:

Answer 1

Explicație pas cu pas:

a)

$x = ( \sqrt{3} + {(3 \sqrt{3})}^{3}) : 7 = ( \sqrt{3} + {3}^{3} \cdot 3 \sqrt{3}) : 7 = ( \sqrt{3} + 81 \sqrt{3}) : 7 = \dfrac{82 \sqrt{3}}{7}$

$y = \Big( \dfrac{ \sqrt{3} }{ \sqrt{2} } - \dfrac{ \sqrt{2} }{ \sqrt{3} } \Big) : \dfrac{1}{ \sqrt{2} } = \Big( \dfrac{ \sqrt{3} }{ \sqrt{2} } - \dfrac{ ^{ \sqrt{3} )} \sqrt{2} }{ \sqrt{3} } \Big) \cdot \sqrt{2} = \sqrt{3} - \dfrac{2 \sqrt{3} }{3} = \dfrac{\sqrt{3} }{3}$

$m_{g} = \sqrt{xy} = \sqrt{\dfrac{82 \sqrt{3}}{7} \cdot \dfrac{\sqrt{3} }{3}} = \sqrt{ \dfrac{82}{7} } = \dfrac{\sqrt{574} }{7}$

b)

$x = ( \sqrt{2} + {(2 \sqrt{2})}^{3}) : 3 = ( \sqrt{2} + {2}^{3} \cdot 2 \sqrt{2}) : 3 = ( \sqrt{2} + 16\sqrt{2}) : 3 = \dfrac{17 \sqrt{2} }{3}$

$y = \bigg[ {\Big( \dfrac{ \sqrt{5} }{ \sqrt{2} } - \dfrac{\sqrt{2} }{ \sqrt{5} } \Big) \cdot \dfrac{ \sqrt{5} }{3}\bigg]}^{3} = \bigg[ {\Big( \dfrac{ \sqrt{10} }{2} - \dfrac{\sqrt{10} }{5} \Big) \cdot \dfrac{ \sqrt{5} }{3}\bigg]}^{3} = {\Big( \dfrac{3\sqrt{10} }{10} \cdot \dfrac{\sqrt{5} }{3} \Big)}^{3} = {\Big(\dfrac{ \sqrt{50} }{10} \Big)}^{3} = {\Big(\dfrac{5 \sqrt{2} }{10} \Big)}^{3} = {\Big(\dfrac{ \sqrt{2} }{2} \Big)}^{3} = \dfrac{ 2\sqrt{2} }{8} = \dfrac{ \sqrt{2} }{4}$

=>

$m_{g} = \sqrt{xy} = \sqrt{\dfrac{17 \sqrt{2} }{3} \cdot \dfrac{ \sqrt{2} }{4}} = \sqrt{\dfrac{17}{6}} = \dfrac{\sqrt{102} }{6}$