Răspuns:
media aritmetică şi media geometrică
Explicație pas cu pas:
a)
[tex]y = \sqrt{14 - 6 \sqrt{5} } = \sqrt{9 + 5 - 6 \sqrt{5} } = \\ = \sqrt{ {3}^{2} - 2 \cdot 3 \cdot\sqrt{5} + {( \sqrt{5} )}^{2} } = \sqrt{ ({3 - \sqrt{5} })^{2} } = |3 - \sqrt{5} | \\ = 3 - \sqrt{5} [/tex]
[tex]m_{a} = \frac{x + y}{2} = \\ = \frac{3 + \sqrt{5} + 3 - \sqrt{5} }{2} = \frac{6}{2} = 3[/tex]
[tex]m_{g} = \sqrt{x\cdot y} = \sqrt{(3 + \sqrt{5} )(3 - \sqrt{5} )} = \\ = \sqrt{ {3}^{2} - {( \sqrt{5} )}^{2} } = \sqrt{9 - 5} = \sqrt{4} = 2[/tex]
b)
[tex]x = \sqrt{ {(3 - 2 \sqrt{3} )}^{2} } = |3 - 2 \sqrt{3} | = 2 \sqrt{3} - 3 \\ [/tex]
[tex]y = \sqrt{21 + 12 \sqrt{3} } = \sqrt{9 + 12 + 12 \sqrt{3}} = \\ = \sqrt{ {3}^{2} + 2 \cdot 3 \cdot 2 \sqrt{3} + {(2 \sqrt{3} )}^{2} } = \sqrt{ {(3 + 2 \sqrt{3} )}^{2} } \\ = |3 + 2 \sqrt{3} | = 2 \sqrt{3} + 3[/tex]
[tex]m_{a} = \frac{x + y}{2} = \frac{2 \sqrt{3} - 3 + 2 \sqrt{3} + 3}{2} = \\ = \frac{4 \sqrt{3} }{2} = 2 \sqrt{3} [/tex]
[tex]m_{g} = \sqrt{xy} = \sqrt{(2 \sqrt{3} - 3)(2 \sqrt{3} + 3)} = \\ = \sqrt{ {(2 \sqrt{3} )}^{2} - {(3)}^{2} } = \sqrt{12 - 9} = \sqrt{3} [/tex]
