Question

Calculați media geometrică a numerelor:
a= √128
b= 10x (1 supra radical din 50 + radical din 5 supra 10 - 1 supra radical din 20)
90 puncte!!!!​

Answer 1

$\it a=\sqrt{128}=\sqrt{64\cdot2}=8\sqrt2\\ \\ \\ b=10\cdot\Big(\dfrac{^{\sqrt2)}1}{\ \ \sqrt{50}} +\dfrac{\sqrt5}{10}-\dfrac{^{\sqrt5)}1}{ \ \ \sqrt{20}}\Big)=10\cdot\Big(\dfrac{\sqrt2}{10}+\dfrac{\sqrt5}{10}-\dfrac{\sqrt5}{10}\Big)=10\cdot\dfrac{\sqrt2}{10}=\sqrt2$

$\it m_g=\sqrt{a\cdot b}=\sqrt{8\sqrt2\cdot\sqrt2}=\sqrt{8\cdot2}=\sqrt{16}=4$

Answer 2

Răspuns:

4

Explicație pas cu pas:

$a = \sqrt{128} = 8 \sqrt{2}$

$b = 10 \cdot \Big( \frac{1}{ \sqrt{50} } + \frac{ \sqrt{5} }{10} - \frac{1}{ \sqrt{20} }\Big) = \\ = 10 \cdot \Big( \frac{^{\sqrt{2})}1}{5\sqrt{2} } + \frac{ \sqrt{5} }{10} - \frac{^{\sqrt{5})}1}{2 \sqrt{5} }\Big)\\ = 10 \cdot \frac{\sqrt{2} + \sqrt{5} - \sqrt{5} }{10} = \sqrt{2}$

$m_{g} = \sqrt{ab} = \sqrt{8 \sqrt{2} \cdot \sqrt{2} } = \sqrt{16} = \bf 4$