Question

Fie a=(4-√3)(√5+ √4) (√9-√8) si b =(√4+ √3)(√5-√4) (√9+√8).
Să se calculeze media geometrică si media aritmetica a celor 2 numere.

Answer 1

Explicație pas cu pas:

$a = ( \sqrt{4} - \sqrt{3})( \sqrt{5} + \sqrt{4} )( \sqrt{9} - \sqrt{8}) = (2 - \sqrt{3})( \sqrt{5} + 2)(3 - 2\sqrt{2}) = (2 \sqrt{5} + 4 - \sqrt{15} - 2 \sqrt{3})(3 - 2 \sqrt{2}) = 6 \sqrt{5} - 4 \sqrt{10} + 12 - 8 \sqrt{2} - 3 \sqrt{15} + 2 \sqrt{30} - 6 \sqrt{3} + 4 \sqrt{6}$

$b = ( \sqrt{4} + \sqrt{3})( \sqrt{5} - \sqrt{4})( \sqrt{9} + \sqrt{8}) = (2 + \sqrt{3})( \sqrt{5} - 2)(3 + 2\sqrt{2}) = (2 \sqrt{5} - 4 + \sqrt{15} - 2 \sqrt{3})(3 + 2 \sqrt{2}) = 6 \sqrt{5} + 4 \sqrt{10} - 12 - 8 \sqrt{2} + 3 \sqrt{15} + 2 \sqrt{30} - 6 \sqrt{3} - 4 \sqrt{6}$

$a + b = 6 \sqrt{5} - 4 \sqrt{10} + 12 - 8 \sqrt{2} - 3 \sqrt{15}+ 2 \sqrt{30} - 6 \sqrt{3} + 4 \sqrt{6} + 6 \sqrt{5} + 4 \sqrt{10} - 12 - 8 \sqrt{2} + 3 \sqrt{15} + 2 \sqrt{30} - 6 \sqrt{3} - 4 \sqrt{6} = 12 \sqrt{5} - 16 \sqrt{2} + 4 \sqrt{30} - 12 \sqrt{3} = 4(3 \sqrt{5} - 4 \sqrt{2} + \sqrt{30} - 3 \sqrt{3} )$

$m_{a} = \frac{a + b}{2} = \frac{4(3 \sqrt{5} - 4 \sqrt{2} + \sqrt{30} - 3 \sqrt{3} )}{2}$

$\bf m_{a} = 2(3 \sqrt{5} - 4 \sqrt{2} + \sqrt{30} - 3 \sqrt{3} )$

$a \cdot b = (2 - \sqrt{3})( \sqrt{5} + 2)(3 - 2\sqrt{2})(2 + \sqrt{3})( \sqrt{5} - 2)(3 + 2\sqrt{2}) = (2 - \sqrt{3})(2 + \sqrt{3})( \sqrt{5} + 2)( \sqrt{5} - 2)(3 - 2\sqrt{2})(3 + 2\sqrt{2}) = (4 - 3)(5 - 4)(9 - 8) = 1 \cdot 1 \cdot 1 = 1$

$\bf m_{g} = \sqrt{a \cdot b} = \sqrt{1} = 1$