Решить биуравнения ! только 2 и 4​

2) (x-2)²(x²-4x)+3=0

(x²-4x-4)(x²-4x)+3 = 0

y = x²-4x

(y-4)y + 3 = 0

y² - 4y + 3 = 0

y1 = 3

y2 = 1

x²-4x = 3

x²-4x-3=0

x1 = 3

x2 = -1

x²-4x = 1

x²-4x-1=0

d1 = 4+1 = 5

x3 = 2+√5

x4 = 2-√5

ответ: 2-√5; -1; 3; 2+√5.

4) (x²-5x+2)(x²-5x-1) = 28

y = x²-5x

(y+2)(y-1) - 28 = 0

y² - y + 2y - 2 - 28 = 0

y² + y - 30 = 0

y1 = -6

y2 = 5

x²-5x = -6

x² - 5x + 6 = 0

x1 = 5

x2 = -1

x²-5x = 5

x² - 5x - 5 = 0

d = 25 + 20 = 45 = (3√5)²

x3 = 2,5 + 1,5√5

x4 = 2,5 - 1,5√5

ответ: 2,5-1,5√5; -1; 5; 2,5+1,5√5.