Question

решите пожалуйста!!!!

Answer 1

а) х² - 10х + 21 ≥ 0

За теоремою Вієта:

х1 + х2 = 10

х1×х2 = 21

х1 = 3, х2 = 7

хЄ(-∞; 3] U [7; +∞)

д) 9х² - 8х - 1 ≤ 0

D = b² - 4ac = 64 + 36 = 100

$x_{1} = \frac{ - b - \sqrt{D} }{2a} = \frac{8 - 10}{18} = - \frac{1}{9}$

$x_{2}= \frac{ - b + \sqrt{D} }{2a} = \frac{ 8 + 10}{18} = 1$

xЄ[-1/9; 1]

и) -7х² + 18х + 9 ≥ 0

D = b² - 4ac = 18² - 4×(-7)×9 = 576

$x_{1} = \frac{ - b - \sqrt{D} }{2a} = \frac{ - 18 - 24}{ - 14} = 3$

$x_{2}= \frac{ - b + \sqrt{D} }{2a} = \frac{ - 18 + 24}{-14} = - \frac{3}{7}$

xЄ[-3/7; 3]