Не вычисляя корней уравнения 2х²-6х+3=0 найдите - вопрос №70585062630 от taaas7107 28.04.2020 02:15

Question

Алгебра: Не вычисляя корней уравнения 2х²-6х+3=0 найдите а) б) в)

taaas7107 · Answer

По теореме Виета x1 + x2 = 6/2 = 3; x1*x2 = 3/2 = 1,5.
a) x1*x2^5 + x1^5*x2 = x1*x2*(x1^4 + x2^4) =
= 1,5*(x1^4 + 2x1^2*x2^2 + x2^4 - 2x1^2*x2^2) =
= 1,5*((x1^2 + x2^2)^2 - 2*(1,5)^2) =
= 1,5*((x1^2 + 2x1*x2 + x2^2 - 2x1*x2)^2) - 2*2,25) =
= 1,5*( [ (x1+x2)^2 - 2*1,5 ]^2 - 4,5) = 1,5*((3^2 - 3)^2 - 4,5) =
= 1,5*(6^2 - 4,5) = 1,5*(36 - 4,5) = 1,5*31,5 = 47,25
b) x1^4 + x2^4 = x1^4 + 2x1^2*x2^2 + x2^4 - 2x1^2*x2^2 =
(x1^2 + x2^2)^2 - 2*(1,5)^2 = (x1^2 + 2x1*x2 + x2^2 - 2x1*x2)^2) - 2*2,25 =
[ (x1+x2)^2 - 2*1,5 ]^2 - 4,5 = (3^2 - 3)^2 - 4,5 = 36 - 4,5 = 31,5
c) $\frac{x1}{x2^2} + \frac{x2}{x1^2} = \frac{x1^3+x2^3}{x1^2*x2^2} = \frac{(x1+x2)(x1^2-x1*x2+x2^2)}{(x1*x2)^2} =$
$= \frac{3(x1^2+2x1x2+x2^2-3x1x2)}{(1,5)^2} = \frac{1,5*2((x1+x2)^2-3*1,5)}{(1,5)^2} = \frac{2(3^2-4,5)}{1,5} = \frac{2*4,5}{1,5}=6$