∛(1+√(x-1))+∛(1-√(x-1))=2

Предыдущий
вопрос Следующий
вопрос

To solve this equation, we can start by letting y = √(x-1), then rewrite the equation in terms of y:

∛(1+y) + ∛(1-y) = 2

Now, we can cube both sides of the equation to eliminate the cube roots:

(∛(1+y) + ∛(1-y))^3 = 2^3

Expanding the left side using the binomial theorem, we get:

(1+y + 1-y + 3(∛(1+y))(∛(1-y))(∛(1+y) + ∛(1-y)) = 8

Simplifying the terms, we have:

2 + 3(∛(1+y))(∛(1-y) = 8

Now, substitute back y = √(x-1) to get:

2 + 3√(1+y)√(1-y) = 8

Now, square both sides to remove the square roots:

(2 + 3√(1+y)√(1-y))^2 = 8^2

Expanding and simplifying the left side, we get:

4 + 12√(1+y)√(1-y) + 9(1+y)(1-y) = 64

Simplify further:

4 + 12√(1+y)√(1-y) + 9(1-y^2) = 64

4 + 12√(1+y)√(1-y) + 9 - 9y^2 = 64

13 + 12√(1+y)√(1-y) - 9y^2 = 64

Then solve this equation for y, and finally solve for x.