Question

Решите систему неравенств:
x'+4x+3<0
(16-3(4-x)>X

Answer 1

Ответ:

5x + 3 < 0

5x < -3

x < -0,6

x € (-∞; -0,6)

52 - 13x > x

-13x - x > 52

-14x > 52

-x > 26/7

x < -(26/7)

x € (-∞; -26/7)

(-∞; -0,6) *пересечение* (-∞; -26/7) = (-∞; -26/7)

x € (-∞; -26/7)

Answer 2

$\displaystyle\bf\\\left \{ {{x^{2} +4x+3 < 0} \atop {16-3(4-x) > x}} \right.\\\\\\\left \{ {{(x-1)\cdot(x-3) < 0} \atop {16-12+3x-x > 0} \right. \\\\\\\left \{ {{(x-1)\cdot(x-3) < 0} \atop {2x > -4}} \right. \\\\\\\left \{ {{(x-1)\cdot(x-3) < 0} \atop {x > -2}} \right.\\\\\\1) \ \ + + + + + \boxed{(1)- - - - - (3)}+ + + + + \\\\2) \ \ \ (-2)///////////////////////////////\\\\\\Otvet \ : \ x\in(1 \ ; \ 3)$