Question

Решите уравнения 1) (y-3)(y-3) =(y-1) ^2 2) 27x^2(1-x)=(1-x)^3

Answer 1

1)

(y-3)(y-3) = (y-1)²,

у² - 9 = у² - 2у + 1,

у² - 9 - у² + 2у - 1 = 0,

2у - 10 = 0,

2у = 10,

у = 5,

2)

27x²*(1 - x) = (1 - x)³,

27х² - 27х³ = 1 - 3х + 3х² - х³,

27х² - 27х³ - 1 + 3х - 3х² + х³ = 0,

27х²*(1 - х) + 3х*(1 - х) + (х³ - 1) = 0,

(1 - х)(27х² + 3х) + (х - 1)(х² + х + 1) = 0,

(1 - х)(27х² + 3х) - (1 - х)(х² + х + 1) = 0,

(1 - х)(27х² + 3х  -  х² - х - 1) = 0,

(1 - х)(26х² + 2х - 1) = 0,

1 - х = 0,   26х² + 2х - 1 = 0,

х1 = 1,      Д = 2² - 4*26*(-1) = 4 + 104 = 108,

               х2 = (-2 + √108) / 2*26 = (-2 + 6√3) / 2*26 = -(1 - 3√3)/26,

              х3 = (-2 - √108) / 2*26 = (-2 - 6√3) / 2*26 = -(1 + 3√3)/26

Answer 2

1. (y-3)(y-3) = (y-1)^2

Expanding the left-hand side, we get:

y^2 - 6y + 9 = y^2 - 2y + 1

Simplifying, we get:

4y = 8

y = 2

Therefore, the solution to the equation is y = 2.

2. 27x^2(1-x) = (1-x)^3

We can simplify the equation by dividing both sides by (1-x):

27x^2 = (1-x)

Expanding the right-hand side, we get:

27x^2 = 1 - 2x + x^2

Bringing all the terms to one side, we get:

26x^2 + 2x - 1 = 0

We can solve this quadratic equation using the quadratic formula:

x = (-2 ± sqrt(2^2 - 4(26)(-1))) / (2(26)) x = (-2 ± sqrt(212)) / 52

Simplifying, we get:

x = (-1 ± sqrt(53)) / 26

Therefore, the solutions to the equation are x = (-1 + sqrt(53)) / 26 and x = (-1 - sqrt(53)) / 26.