Вопрос задан 24.07.2023 в 00:43. Предмет Алгебра. Спрашивает Аликина Оля.

1) x³ - 64x > 0 2) x³ ≤ 2x 3) x + ≤ 6 4) x + ≥ 3

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Отвечает Сандалов Ваня.
Аллалвоовопоклдвлалаща

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Отвечает нейросеть ChatGpt. Будьте внимательны, ответы нейросети могут содержать ошибочные или выдуманные данные.

Let's solve each inequality step by step:

  1. x³ - 64x > 0

To solve this inequality, first factor out x from the expression:

x(x² - 64) > 0

Now, factor the expression further:

x(x + 8)(x - 8) > 0

Next, find the critical points by setting each factor equal to zero:

x = 0, x + 8 = 0, x - 8 = 0

Solving for x:

x = 0, x = -8, x = 8

Now, we can use test intervals to determine the sign of the expression in each interval. Choose test points from each interval and evaluate the expression:

Test point: x = -10

(-10)(-10 + 8)(-10 - 8) = -10(-2)(-18) = 360

The expression is positive in this interval.

Test point: x = 5

(5)(5 + 8)(5 - 8) = 5(13)(-3) = -195

The expression is negative in this interval.

Test point: x = 10

(10)(10 + 8)(10 - 8) = 10(18)(2) = 360

The expression is positive in this interval.

Now, we can write the solution by considering where the expression is greater than zero:

Solution: x < -8 or 0 < x < 8

  1. x³ ≤ 2x

To solve this inequality, first, move all terms to one side of the inequality:

x³ - 2x ≤ 0

Next, factor out x:

x(x² - 2) ≤ 0

Now, find the critical points by setting each factor equal to zero:

x = 0, x² - 2 = 0

Solving for x:

x = 0, x = ±√2

Now, use test intervals to determine the sign of the expression in each interval:

Test point: x = -1

(-1)((-1)² - 2) = -1(1 - 2) = 1

The expression is positive in this interval.

Test point: x = 1.5

(1.5)((1.5)² - 2) = 1.5(1.25) = 1.875

The expression is positive in this interval.

Test point: x = 3

(3)((3)² - 2) = 3(7) = 21

The expression is positive in this interval.

Now, write the solution by considering where the expression is less than or equal to zero:

Solution: -√2 ≤ x ≤ 0

  1. x + 2 ≤ 6

To solve this inequality, first, move the constant term to the other side of the inequality:

x + 2 - 2 ≤ 6 - 2

Simplify:

x ≤ 4

Solution: x ≤ 4

  1. x + 1 ≥ 3

To solve this inequality, first, move the constant term to the other side of the inequality:

x + 1 - 1 ≥ 3 - 1

Simplify:

x ≥ 2

Solution: x ≥ 2

To summarize the solutions:

  1. x < -8 or 0 < x < 8
  2. -√2 ≤ x ≤ 0
  3. x ≤ 4
  4. x ≥ 2
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