Question

calculati media aritmetica si media geometrica a numerelor :
$2 \sqrt{ 3 } + 3$
si
$\sqrt{21 + 12 \sqrt{3} }$

Answer 1

Răspuns:

3+2√3

3+2√3

Explicație pas cu pas:

21+12√3=9+2*6√3+12=3²+2*3*2√3+(2√3)²=(3+2√3)²

atunci √(3+2√3)²=|3+2√3|=3+2√3

cele 2 numere fiind egale, m.a.=m.g=3+2√3

Answer 2

Fie x,y-nr.

$x = 2 \sqrt{3} + 3$

$y = \sqrt{21 + 12 \sqrt{3} }$

Formula radicalilor compuşi :

$\sqrt{a \pm \sqrt{b} } = \sqrt{ \frac{ a + \sqrt{ {a}^{2} - b} }{2} } \pm\sqrt{ \frac{ a - \sqrt{ {a}^{2} - b} }{2} }$

$y = \sqrt{21 + 12 \sqrt{3} }$

$y = \sqrt{21 + \sqrt{ {12}^{2} \times 3 } }$

$y = \sqrt{21 + \sqrt{144 \times 3} }$

$y = \sqrt{21 + \sqrt{432} }$

$y = \sqrt{ \frac{ 21 + \sqrt{ {21}^{2} - 432 } }{2} } + \sqrt{ \frac{ 21 - \sqrt{ {21}^{2} - 432 } }{2} }$

$y = \sqrt{ \frac{21 + \sqrt{441 - 432} }{2} } + \sqrt{ \frac{21 - \sqrt{441 - 432} }{2} }$

$y = \sqrt{ \frac{21 + \sqrt{9} }{2} } + \sqrt{ \frac{21 - \sqrt{9} }{2} }$

$y = \sqrt{ \frac{21 + 3}{2} } + \sqrt{ \frac{21 - 3}{2} }$

$y = \sqrt{ \frac{24}{2} } + \sqrt{ \frac{18}{2} }$

$y = \sqrt{12} + \sqrt{9}$

$y = 2 \sqrt{3} + 3$

$m_{a} = \frac{x + y}{2}$

$m_{a} = \frac{2 \sqrt{3} + 3 + 2 \sqrt{3} + 3 }{2}$

$m_{a} = \frac{ 4\sqrt{3} + 6 }{2}$

$m_{a} = \frac{2(2 \sqrt{3} + 3) }{2}$

$m_{a} = 2 \sqrt{3} + 3$

$m_{g} = \sqrt{xy}$

$m_{g} = \sqrt{(2 \sqrt{3} + 3)(2 \sqrt{3} + 3) }$

$m_{g} = \sqrt{ {(2 \sqrt{3} + 3) }^{2} }$

$m_{g} = {(2 \sqrt{3} + 3) }^{ \frac{2}{2} }$

$m_{g} = {(2 \sqrt{3} + 3) }^{1}$

$m_{g} = 2 \sqrt{3} + 3$