((2x^2+4)(3x-x^2))/(2x+5)^3<=0 (x^2-2x-1)/((2x-5(x+2)^2)<0

Предыдущий
вопрос Следующий
вопрос

To solve the inequality

(2x^2 + 4)(3x - x^2)/(2x + 5)^3 <= 0,

we must first find the critical points where the expression is undefined or changes sign.

Setting the denominator of the fraction equal to 0, we find that the critical point is x = -5/2.

Next, we set the numerator equal to 0 to find additional critical points.

2x^2 + 4 = 0
2x^2 = -4
x^2 = -2
This equation has no real solutions, so there are no additional critical points.

Therefore, the critical points for the first inequality are x = -5/2.

We can test the intervals between the critical points and the endpoints (-∞, -5/2), (-5/2, ∞) to determine where the inequality holds true.

For the interval (-∞, -5/2):
We choose a test point, such as x = -3. Plugging that into the expression gives

(2(-3)^2 + 4)(3(-3) - (-3)^2)/(2(-3) + 5)^3 = (4)(-15 + 9)/(8)^3 < 0

Since the expression is negative in this interval, it satisfies the inequality.

For the interval (-5/2, ∞):
We choose a test point, such as x = 0. Plugging that into the expression gives

(2(0)^2 + 4)(3(0) - 0^2)/(2(0) + 5)^3 = (4)(0)/(5)^3 = 0

Since the expression is equal to 0 in this interval, it does not satisfy the inequality.

Therefore, the solution to the inequality (2x^2 + 4)(3x - x^2)/(2x + 5)^3 <= 0 is x ∈ (-∞, -5/2].

For the second inequality

(x^2 - 2x - 1)/((2x - 5)(x + 2)^2) < 0

We first find the critical points where the expression is undefined or changes sign.

Setting the denominator of the fraction equal to 0, we find that the critical points are x = 5/2, -2.

Next, we set the numerator equal to 0 to find additional critical points.

x^2 - 2x - 1 = 0
This quadratic has solutions x = 1 - √2 and x = 1 + √2.

Therefore, the critical points for the second inequality are x = 1 - √2, 1 + √2, 5/2, -2.

We can test the intervals between the critical points and the endpoints to determine where the inequality holds true.

Solving for each interval, we find that the solution to the inequality (x^2 - 2x - 1)/((2x - 5)(x + 2)^2) < 0 is -2 < x < 1 - √2 or 1 + √2 < x < 5/2.