Question

Calculați media aritmetică și geometrică a numerelor reale a și b. a=6(4supra√3 + 6 supra √27) - (2√12+ 6supra√3).
b= (3√2supra2 - 1supra3√2)x 4√6 - (4supra√8 - 2supra3√2) x 2√6​

Answer 1

$\displaystyle a=6\Bigg(\frac{4}{\sqrt{3} }+\frac{6}{\sqrt{27} } \Bigg)-\Bigg(2\sqrt{12}+\frac{6}{\sqrt{3} }\Bigg)=\\\\=6\Bigg(\frac{4}{\sqrt{3} }+\frac{6}{3\sqrt{3} }\Bigg)-\Bigg(2 \cdot 2\sqrt{3} +\frac{6}{\sqrt{3} }\Bigg)=6\Bigg(\frac{4}{\sqrt{3} }+\frac{6}{3\sqrt{3} }\Bigg)-\Bigg(4\sqrt{3} +\frac{6}{\sqrt{3} }\Bigg)=\\\\=6\cdot \frac{12+6}{3\sqrt{3} }-\frac{12+6}{\sqrt{3}}=2 \cdot \frac{18}{\sqrt{3} } -\frac{18}{\sqrt{3} }=\frac{36-18}{\sqrt{3} }=\frac{18}{\sqrt{3} } =\frac{18\sqrt{3} }{3}=6\sqrt{3}$

$\displaystyle b=\Bigg(\frac{3\sqrt{2} }{2} -\frac{1}{3\sqrt{2} }\Bigg)\cdot 4\sqrt{6} -\Bigg(\frac{4}{\sqrt{8} } -\frac{2}{3\sqrt{2} } \Bigg) \cdot 2\sqrt{6} =\\\\=\frac{18-2 }{6\sqrt{2} }\cdot 4\sqrt{6} -\Bigg(\frac{4}{2\sqrt{2} }-\frac{2}{3\sqrt{2} } \Bigg) \cdot 2\sqrt{6} = \frac{16 }{6\sqrt{2} }\cdot 4\sqrt{6} -\frac{12-4}{6\sqrt{2} } \cdot2\sqrt{6} =\\\\ =\frac{64\sqrt{6} -16\sqrt{6} }{6\sqrt{2}}= \frac{48\sqrt{6} }{6\sqrt{2} }= 8\sqrt{3}$

$\displaystyle m_a=\frac{a+b}{2} =\frac{6\sqrt{3} +8\sqrt{3} }{2} =\frac{14\sqrt{3} }{2} =7\sqrt{3} \\\\ m_g=\sqrt{a \cdot b}=\sqrt{6\sqrt{3} \cdot 8\sqrt{3} } =\sqrt{48 \cdot 3}= \sqrt{144}=12$