Задача 58689 ...

602a90393cfdeb46290c929b

20.03.2021 23:10:02

Решите неравенство

1/log(x-3)0,5 - log(2 - x)(x + 5) + log(0,5)(x + 5) ≥ log(x - 2)(x - 3)

5f3ea7e3faf909182968ddd9

21.03.2021 12:11:58

★

ОДЗ:
[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]