Question

(2x-1)²-(x-1)(x+7)</=5 ​

Answer 1

Ответ:

Ответ смотри во вложении

Answer 2

To solve the inequality (2x-1)² - (x-1)(x+7) ≤ 5, you can follow these steps:

1. Expand and simplify the left side of the inequality: (2x-1)² - (x-1)(x+7) ≤ 5 (4x² - 4x + 1) - (x² + 6x - 7) ≤ 5

2. Distribute the negative sign to the terms inside the second parenthesis: 4x² - 4x + 1 - x² - 6x + 7 ≤ 5

3. Combine like terms on both sides of the inequality: (4x² - x²) + (-4x - 6x) + (1 + 7) ≤ 5

4. Simplify further: 3x² - 10x + 8 ≤ 5

5. Subtract 5 from both sides of the inequality: 3x² - 10x + 8 - 5 ≤ 0

6. Simplify the inequality: 3x² - 10x + 3 ≤ 0

Now, we need to find the values of x that satisfy this quadratic inequality. To do this, we can factor the quadratic expression:

3x² - 10x + 3 = 0

You can factor this quadratic equation as follows:

(3x - 1)(x - 3) = 0

Now, we have two critical values where the inequality might change: x = 1/3 and x = 3. We will use these critical values to test the intervals:

1. Test the interval x < 1/3: Choose a test point, let's say x = 0, and plug it into the inequality: 3(0)² - 10(0) + 3 ≤ 0 3 ≤ 0 This is not true, so the inequality is not satisfied in this interval.

2. Test the interval 1/3 < x < 3: Choose a test point, let's say x = 2, and plug it into the inequality: 3(2)² - 10(2) + 3 ≤ 0 12 - 20 + 3 ≤ 0 -5 ≤ 0 This is true, so the inequality is satisfied in this interval.

3. Test the interval x > 3: Choose a test point, let's say x = 4, and plug it into the inequality: 3(4)² - 10(4) + 3 ≤ 0 48 - 40 + 3 ≤ 0 11 ≤ 0 This is not true, so the inequality is not satisfied in this interval.

So, the solution to the inequality 3x² - 10x + 3 ≤ 0 is: 1/3 ≤ x ≤ 3

Therefore, the values of x that satisfy the original inequality (2x-1)² - (x-1)(x+7) ≤ 5 are: 1/3 ≤ x ≤ 3