Question

clasa a10a
4) 1. a,f,g​

Answer 1

Explicație pas cu pas:

$S = x_{1} + x_{2} = -3$

$P = x_{1} x_{2} = 3$

a)

$x_{1}^{2} + x_{2}^{2} = (x_{1} + x_{2})^{2} - 2x_{1} x_{2} = \\ = S^{2} - 2P = (-3)^{2} - 2 \cdot 3 = 9 - 6 = 3$

f)

$\dfrac{x_{1}}{x_{2}} + \dfrac{x_{2}}{x_{1}} = \dfrac{x_{1}^{2} + x_{2}^{2}}{x_{1} x_{2}} = \dfrac{S^{2} - 2P}{P} = \dfrac{3}{3} = 1$

g)

$\dfrac{x_{1}^{2}}{x_{2}} + \dfrac{x_{2}^{2}}{x_{1}} = \dfrac{x_{1}^{3} + x_{2}^{3}}{x_{1} x_{2}} = \dfrac{(x_{1} + x_{2})^{3} - 3 x_{1} x_{2}(x_{1} + x_{2})}{x_{1} x_{2}} = \\ = \dfrac{S^{3} - 3SP}{P} = \dfrac{( - 3)^{3} - 3 \cdot ( - 3) \cdot 3}{3} = \dfrac{ - 27 + 27}{3} = 0$