Question

2.12. 1) (х-2)2-49=0;2) 9(2x+3)? -25 - 0;3) 2(3x+5)? = 7(3x+5);4) (3x-1) 4-12х.​

Answer 1

Пошаговое объяснение:

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Answer 2

Let's solve each of these equations step by step:

1. (x - 2)^2 - 49 = 0:

First, expand the square:

(x^2 - 4x + 4) - 49 = 0

Now, combine like terms:

x^2 - 4x - 45 = 0

This is a quadratic equation. We can solve it by factoring or using the quadratic formula. Factoring, we get:

(x - 9)(x + 5) = 0

Now, set each factor equal to zero:

x - 9 = 0 or x + 5 = 0

Solve for x in each equation:

x = 9 or x = -5

So, there are two solutions: x = 9 and x = -5.

2. 9(2x + 3)^2 - 25 = 0:

First, divide both sides by 9:

(2x + 3)^2 - 25/9 = 0

Now, take the square root of both sides:

2x + 3 = ±√(25/9)

2x + 3 = ±(5/3)

Now, solve for x in each case:

Case 1: 2x + 3 = 5/3

Subtract 3 from both sides:

2x = 5/3 - 3

2x = 5/3 - 9/3

2x = -4/3

Divide by 2:

x = (-4/3) / 2

x = -2/3

Case 2: 2x + 3 = -5/3

Subtract 3 from both sides:

2x = -5/3 - 3

2x = -5/3 - 9/3

2x = -14/3

Divide by 2:

x = (-14/3) / 2

x = -7/3

So, there are two solutions: x = -2/3 and x = -7/3.

3. 2(3x + 5)^2 = 7(3x + 5):

First, divide both sides by 2:

(3x + 5)^2 = (7/2)(3x + 5)

Now, take the square root of both sides:

3x + 5 = ±√((7/2)(3x + 5))

3x + 5 = ±√(21x/2 + 35/2)

Now, we'll solve for x in each case:

Case 1: 3x + 5 = √(21x/2 + 35/2)

Subtract 5 from both sides:

3x = √(21x/2 + 35/2) - 5

Square both sides to eliminate the square root:

9x^2 = 21x/2 + 35/2

Multiply both sides by 2 to eliminate the fraction:

18x^2 = 21x + 35

Rearrange and set the equation equal to zero:

18x^2 - 21x - 35 = 0

Now, you can solve this quadratic equation using the quadratic formula or factoring.

4. (3x - 1)^4 - 12x:

This equation appears to be incomplete. If you have a specific operation or equation you want to solve for, please provide more information, and I'll be happy to help.