Question

lg(x-2)+lg(27-x)<2
Pamagite​

Answer 1

Ответ:

ОДЗ:

$\left \{ {{x - 2 > 0} \atop {27 - x > 0} } \right. \\ \\ \left \{ {{x > 2} \atop {x < 27} } \right. \\ \\ x\in(2;27)$

$lg(x - 2) + lg(27 - x) < 2 \\ lg(x - 2)(27 - x) < 2 \\ (x - 2)(27 - x) < {10}^{2} \\ 27x - {x}^{2} - 54 + 2x - 100 < 0 \\ - {x}^{2} + 29x - 154 < 0 \\ {x}^{2} - 29x + 154 > 0 \\ D= 841 - 616 = 225 \\ x_1 = \frac{29 + 15}{2} = \frac{44}{2} = 22 \\ x_2 = \frac{14}{2} = 7 \\ + \: \: \: \: \: \: - \: \: \: \: \: \: \: \: + \\ - -7 - - 22 - - > \\ x\in( - \infty; 7)U(22; + \infty )$

Пересекаем с ОДЗ

Ответ:

$x\in(2;7)U(22 ;27)$