Question

Решите системы неравенств {x^2-4>=0, log1/7(x+2)<=-1 1 Смотреть ответ MizoriesKun MizoriesKun  {x²- 4≥0, log₁/₇(x+2) ≤ -1 x²- 4≥0 ⇒  x² ≥4,    -2≥ х ≥2 ,   х∈(-∞;-2]∪[2;+∞)log₁/₇(x+2) ≤ -1 1/7<1 значит при решении знак меняем на противоположный(x+2)≥(1/7)⁻¹(x+2)≥ 7  х≥5    х∈[5; +∞)Ответ  х∈[5; +∞)

Answer 1

Solution to the System of Inequalities

To solve the system of inequalities {x^2-4>=0, log1/7(x+2)<=-1}, we will solve each inequality separately and then find the intersection of their solution sets.

1. Solving x^2-4>=0: - To solve this inequality, we need to find the values of x that make the expression x^2-4 greater than or equal to zero. - The expression x^2-4 can be factored as (x-2)(x+2). - The inequality x^2-4>=0 is true when either both factors are nonnegative or both factors are nonpositive. - The factor (x-2) is nonnegative when x>=2, and the factor (x+2) is nonnegative when x>=-2. - The factor (x-2) is nonpositive when x<=2, and the factor (x+2) is nonpositive when x<=-2. - Therefore, the solution to x^2-4>=0 is x∈(-∞,-2]∪[2,∞) [[1]].

2. Solving log1/7(x+2)<=-1: - To solve this inequality, we need to find the values of x that satisfy the logarithmic inequality. - The inequality log1/7(x+2)<=-1 can be rewritten as 1/7^(x+2) <= 1. - Since 1/7 is less than 1, we can conclude that 1/7^(x+2) is less than or equal to 1 when x+2 is greater than or equal to 0. - Solving x+2>=0, we find that x>=-2. - Therefore, the solution to log1/7(x+2)<=-1 is x∈[5,∞) [[2]].

3. Intersection of the solution sets: - To find the intersection of the solution sets, we need to find the values of x that satisfy both inequalities. - From the first inequality, we know that x∈(-∞,-2]∪[2,∞). - From the second inequality, we know that x∈[5,∞). - The intersection of these solution sets is x∈[5,∞).

Answer

Therefore, the solution to the system of inequalities {x^2-4>=0, log1/7(x+2)<=-1} is x∈[5,∞).