Question

8-(х-3)(х+3)=10-(х-1)² ,(2х+1)²-(2х-3)²=4(7х-5)

Answer 1

Вот решение приложено

Answer 2

$8 - (x - 3)( x + 3) = 10 - (x - 1 {)}^{2} \\ 8 - {x}^{2} + 9 = 10 - {x}^{2} + 2x - 1 \\ 8 + 9 = 10 + 2x - 1 \\ 17 = 9 + {2}x \\ - 2x = 9 - 17 \\ - 2x = - 8 \\ x = 4$

$(2x + 1 {)}^{2} - (2x - 3 {)}^{2} = 4(7x - 5 )\\ 4 {x}^{2} + 4x + 1 - (4 {x}^{2} - 12x + 9) = 28x - 20 \\ 4 {x}^{2} + 4x + 1 - 4 {x}^{2} + 12x - 9 = 28x - 20 \\ 16x - 8 = 28x - 20 \\ 16x - 28x = - 20 + 8 \\ - 12x = - 12 \\ x = 1$

Answer 3

To solve the given equations, let's work on them one by one.

Equation 1: 8 - (x - 3)(x + 3) = 10 - (x - 1)²

Expanding the terms and simplifying, we have:

8 - (x² - 3x + 3x - 9) = 10 - (x² - 2x + 1)

8 - (x² - 9) = 10 - x² + 2x - 1

8 - x² + 9 = 10 - x² + 2x - 1

-x² + 2x + 17 = -x² + 2x + 9

Subtracting (-x² + 2x) from both sides:

17 = 9

The equation 17 = 9 is not true, which means there is no solution to the equation. Therefore, Equation 1 has no solutions.

Moving on to Equation 2: (2x + 1)² - (2x - 3)² = 4(7x - 5)

Expanding and simplifying the equation:

(4x² + 4x + 1) - (4x² - 12x + 9) = 28x - 20

4x² + 4x + 1 - 4x² + 12x - 9 = 28x - 20

(4x² - 4x²) + (4x + 12x) + (1 - 9) = 28x - 20

16x - 8 = 28x - 20

Subtracting 16x from both sides:

-8 = 12x - 20

Adding 20 to both sides:

12 = 12x

Dividing both sides by 12:

x = 1

Therefore, the solution to Equation 2 is x = 1.