Question

1) -3x^2+x+24>0 2) x^2-2x+1>или равно0 3) 3x^2-4x-7>или равно0 4) (2x-4) (3x+6)<0 5) (2x-7) ( 4x+5) (x-4)<0

Answer 1

Решение задания смотри на фотографии

Answer 2

Let's solve each inequality step by step:

1. -3x^2 + x + 24 > 0

First, find the critical points by setting the expression equal to zero:

-3x^2 + x + 24 = 0

Now, use the quadratic formula to find the roots:

x = (-b ± √(b^2 - 4ac)) / 2a

where a = -3, b = 1, and c = 24.

x = (-(1) ± √(1^2 - 4*(-3)24)) / 2(-3)

x = (-(1) ± √(1 + 288)) / (-6)

x = (-(1) ± √289) / (-6)

x = (-(1) ± 17) / (-6)

So the critical points are x = 3 and x = -4/3.

Now, check the intervals between these critical points and beyond to determine the solution:

1. When x < -4/3: Substitute x = -2 into the inequality: -3(-2)^2 + (-2) + 24 > 0 -12 + (-2) + 24 > 0 10 > 0

2. When -4/3 < x < 3: Substitute x = 0 into the inequality: -3(0)^2 + (0) + 24 > 0 24 > 0

3. When x > 3: Substitute x = 4 into the inequality: -3(4)^2 + (4) + 24 > 0 -48 + 4 + 24 > 0 -20 > 0

The solution is: x < -4/3 or 3 < x.

2. x^2 - 2x + 1 ≥ 0

This is a quadratic equation in standard form. To solve it, factor it if possible:

(x - 1)^2 ≥ 0

The square of any real number is non-negative, so the inequality holds true for all real values of x.

The solution is: x can be any real number.

3. 3x^2 - 4x - 7 ≥ 0

This is a quadratic inequality. We can use the quadratic formula to find the critical points:

x = (-b ± √(b^2 - 4ac)) / 2a

where a = 3, b = -4, and c = -7.

x = (4 ± √((-4)^2 - 4 * 3 * (-7))) / (2 * 3)

x = (4 ± √(16 + 84)) / 6

x = (4 ± √100) / 6

x = (4 ± 10) / 6

So the critical points are x = 2 and x = -1.

Now, check the intervals between these critical points and beyond to determine the solution:

1. When x < -1: Substitute x = -2 into the inequality: 3(-2)^2 - 4(-2) - 7 ≥ 0 12 + 8 - 7 ≥ 0 13 ≥ 0

2. When -1 < x < 2: Substitute x = 0 into the inequality: 3(0)^2 - 4(0) - 7 ≥ 0 -7 ≥ 0

3. When x > 2: Substitute x = 3 into the inequality: 3(3)^2 - 4(3) - 7 ≥ 0 27 - 12 - 7 ≥ 0 8 ≥ 0

The solution is: x ≤ -1 or x ≥ 2.

4. (2x - 4)(3x + 6) < 0

To solve this inequality, find the critical points by setting each factor equal to zero:

2x - 4 = 0 2x = 4 x = 2

3x + 6 = 0 3x = -6 x = -2

Now, check the intervals between these critical points and beyond to determine the solution:

1. When x < -2: Substitute x = -3 into the inequality: (2(-3) - 4)(3(-3) + 6) < 0 (-10)(-3) < 0 30 < 0 (False)

2. When -2 < x < 2: Substitute x = 0 into the inequality: (2(0) - 4)(3(0) + 6) < 0 (-4)(6) < 0 -24 < 0 (True)

3. When x > 2: Substitute x = 3 into the inequality: (2(3) - 4)(3(3) + 6) < 0 (2)(15) < 0 30 < 0 (False)

The solution is: -2 < x < 2.

5. (2x - 7)(4x + 5)(x - 4) < 0

To solve this inequality, find the critical points by setting each factor equal to zero:

2x - 7 = 0 2x = 7 x = 7/2

4x + 5 = 0 4x = -5 x = -5/4

x - 4 = 0 x = 4

Now, check the intervals between these critical points and beyond to determine the solution:

1. When x < -5/4: Substitute x = -2 into the inequality: (2(-2) - 7)(4(-2) + 5)(-2 - 4) < 0 (-11)(-3)(-6) < 0 198 < 0 (False)

2. When -5/4 < x < 4: Substitute x = 0 into the inequality: (2(0) - 7)(4(0) + 5)(0 - 4) < 0 (-7)(5)(-4) < 0 140 < 0 (False)

3. When 4 < x < 7/2: Substitute x = 5 into the inequality: (2(5) - 7)(4(5) + 5)(5 - 4) < 0 (3)(25)(1) < 0 75 < 0 (False)

4. When x > 7/2: Substitute x = 8 into the inequality: (2(8) - 7)(4(8) + 5)(8 - 4) < 0 (9)(37)(4) < 0 1332 < 0 (False)

The solution is: -5/4 < x < 4.

So, to recap:

1. -∞ < x < -4/3 or 3 < x < ∞

2. All real values of x.

3. -∞ < x ≤ -1 or x ≥ 2.

4. -2 < x < 2.

5. -5/4 < x < 4.