Question

Calculaţi media geometrică a numerelor (4√18-3√/50+6√72-√162)×√3 şi 6√24-4√54 +2√486+ √216​

Answer 1

$(4 \sqrt{18} - 3 \sqrt{50} + 6 \sqrt{72} - \sqrt{162}) \cdot \sqrt{3} = (4 \cdot 3 \sqrt{2} - 3 \cdot 5 \sqrt{2} + 6 \cdot 6 \sqrt{2} - 9 \sqrt{2}) \cdot \sqrt{3} = (12 \sqrt{2} - 15 \sqrt{2} + 36 \sqrt{2} - 9 \sqrt{2}) \cdot \sqrt{3} = 24 \sqrt{2} \cdot \sqrt{3} = 24 \sqrt{6}$

$6 \sqrt{24} - 4 \sqrt{54} + 2 \sqrt{486} + \sqrt{216} = 6 \cdot 2 \sqrt{4} - 4 \cdot 3 \sqrt{6} + 2 \cdot 9 \sqrt{6} + 6 \sqrt{6} = 12 \sqrt{4} - 12 \sqrt{6} + 18 \sqrt{6} + 6 \sqrt{6} = 24 \sqrt{6}$

$m_{g} = \sqrt{24 \sqrt{6} \cdot 24 \sqrt{6} } = \sqrt{(24 \sqrt{6})^{2} } = \bf 24 \sqrt{6}$