Помогите с решением: з

1) (3x+2) ^2 = 10+3 (x-2) (x+2)

2) (2x-3) ^2 = 9-2 (x-3) (x+3)

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

4) (x-1) (x^2+x+1) - x^2 (x-1) = 0

1) (3x + 2) (3x + 2) = 10 + 3 (x - 2) (x + 2)

9x² + 12x + 4 = 10 + 3 (x² - 4)

9x² + 12x + 4 = 10 + 3x² - 12

9x² - 3x² + 12x + 4 + 2 = 0

6x² + 12x + 6 = 0

x² + 2x + 1 = 0

D = b² - 4ac = 4 - 4 * 1 = 0 - имеет один корень.

x = - b/2a

x = - 2 / 2 = - 1

Ответ: x = - 1.

2) (2x - 3) (2x - 3) = 9 - 2 (x - 3) (x + 3)

4x² - 12x + 9 = 9 - 2 (x² - 9)

4x² - 12x + 9 = 9 - 2x² + 18

4x² + 2x² - 12x + 9 - 9 - 18 = 0

6x² - 12x - 18 = 0

x² - 2x - 3 = 0

D = b² - 4ac = 4 - 4 * (- 3) = 4 + 12 = 16 = 4²

x1 = (2 + 4) / 2 = 3

x2 = (2 - 4) / 2 = - 1.

3) (x + 2) (x² - 2x + 4) - x² (x + 2) = 0

x³ - 2x² + 4x + 2x² - 4x + 8 - x³ - 2x² = 0

- 2x² + 8 = 0

- 2x² = - 8

2x² = 8

x² = 4

x = 2

x = - 2

4) (x - 1) (x² + x + 1) - x² (x - 1) = 0

x³ + x² + x - x² - x - 1 - x³ + x² = 0

x² - 1 = 0

x² = 1

x = 1

x = - 1

Ответ: x = 1, x = - 1.