Question

Нужен ответ только на 3 задание

Answer 1

$\displaystyle\bf\\\left \{ {{x - 5 \geqslant 0} \atop { {x}^{2} - 2 x- 15 < 0}} \right. \\ \\ 1) \: x - 5 \geqslant 0 \\ x \geqslant 5 \\ \\ 2) \: {x}^{2} - 2x - 15 < 0 \\ \\ {x}^{2} - 2x - 15 = 0 \\ po \: \: \: teoreme \: \: \: vieta \\ {x}^{2} + bx + c = 0\\ x_{1} + x_{2} = - b\\ x_{1} x_{2} = c \\ \\ x_{1} + x_{2} =2 \\ x_{1} x_{2} = - 15 \\ x_{1} = -3 \\ x_{2} = 5\\ \\ {ax}^{2} + bx + c = a(x - x_{1})(x - x_{2}) \\ {x}^{2} - 2x - 15 = (x + 3)(x - 5) \\ \\ (x + 3)(x - 5) < 0 \\ + + + ( - 3) - - - (5) + + + \\ - 3 < x < 5 \\ \displaystyle\bf\\3) \: \left \{ {{x \geqslant 5} \atop { - 3 < x < 5 }} \right. \\ \\ x \: \epsilon\: \varnothing$