Question

X^2-10x+25/x^2-4x-12≥0

Answer 1

Вот списывай на здоровье

Answer 2

To solve the inequality:

$\frac{x^2 - 10x + 25}{x^2 - 4x - 12} \geq 0$

We can start by finding the critical points and determining the sign of the expression in each interval.

First, let's factor the numerator and denominator:

$\frac{(x - 5)^2}{(x - 6)(x + 2)} \geq 0$

The critical points occur when the numerator and denominator equal zero:

Numerator: $x - 5 = 0 \Rightarrow x = 5$

Denominator: $x - 6 = 0 \Rightarrow x = 6$ $x + 2 = 0 \Rightarrow x = -2$

Now, we create a sign chart with these critical points:

Interval (-∞, -2): In this interval, we choose a value less than -2, let's say -3, and substitute it into the expression: $\frac{(-3 - 5)^2}{(-3 - 6)(-3 + 2)} = \frac{64}{27} > 0$ The expression is positive in this interval.

Interval (-2, 5): In this interval, we choose a value between -2 and 5, let's say 0, and substitute it into the expression: $\frac{(0 - 5)^2}{(0 - 6)(0 + 2)} = \frac{25}{-12} < 0$ The expression is negative in this interval.

Interval (5, 6): In this interval, we choose a value between 5 and 6, let's say 5.5, and substitute it into the expression: $\frac{(5.5 - 5)^2}{(5.5 - 6)(5.5 + 2)} = \frac{0.25}{-6.5} < 0$ The expression is negative in this interval.

Interval (6, ∞): In this interval, we choose a value greater than 6, let's say 7, and substitute it into the expression: $\frac{(7 - 5)^2}{(7 - 6)(7 + 2)} = \frac{4}{27} > 0$ The expression is positive in this interval.

Now, we can determine the solution to the inequality based on the sign chart:

$\frac{x^2 - 10x + 25}{x^2 - 4x - 12} \geq 0$

The solution is:

$x \in (-\infty, -2] \cup (5, 6] \cup [6, \infty)$

This means that the inequality is satisfied when x is less than or equal to -2, or when x is greater than 5 and less than or equal to 6, or when x is greater than or equal to 6.