Задача 58689 Задание 15 егэ. Решите пожалуйста...
Условие
Решение
[m]\left\{\begin {matrix}x-3>0, x-3 ≠1 \\x-2>0, x-2 ≠1\\x+5 >0\end {matrix}\right.[/m] ⇒ [m]\left\{\begin {matrix}x>3, x ≠4 \\x>2, x ≠3\\x>-5\end {matrix}\right.[/m]
x ∈ (3;4)U(4;+ ∞)
Переходим к основанию 0,5
[m]\frac{1}{log_{x-3}0,5}=log_{0,5}(x-3)[/m]
[m]log_{x-2}(x+5)=\frac{log_{0,5}(x+5)}{log_{0,5}(x-2)}[/m]
[m]log_{x-2}(x-3)=\frac{log_{0,5}(x-3)}{log_{0,5}(x-2)}[/m]
Уравнение принимает вид:
[m]log_{0,5}(x-3)-\frac{log_{0,5}(x+5)}{log_{0,5}(x-2)}+log_{0,5}(x+5) ≥\frac{log_{0,5}(x-3)}{log_{0,5}(x-2)} [/m]
[m]log_{0,5}(x-3)+log_{0,5}(x+5) ≥\frac{log_{0,5}(x-3)}{log_{0,5}(x-2)}+\frac{log_{0,5}(x+5)}{log_{0,5}(x-2)} [/m]
[m]log_{0,5}(x-3)+log_{0,5}(x+5) ≥\frac{log_{0,5}(x-3)+log_{0,5}(x+5)}{log_{0,5}(x-2)} [/m]
Заменим сумму логарифмов логарифмом произведения
[m]log_{0,5}(x-3)(x+5) ≥\frac{log_{0,5}(x-3)(x+5)}{log_{0,5}(x-2)} [/m]
[m]log_{0,5}(x-3)(x+5) -\frac{log_{0,5}(x-3)(x+5)}{log_{0,5}(x-2)}≥0 [/m]
[m]log_{0,5}(x-3)(x+5)\cdot (1-\frac{1}{log_{0,5}(x-2)})≥0 [/m]