Question

решите уравнение пожалуйста

Answer 1

4 + ⁵√(64y²) = ⁵√(128y⁴)

4 + ⁵√2⁶*⁵√y² = ⁵√2⁷*⁵√y⁴

⁵√y² = x (>=0)

2*⁵√2²*x² - 2*⁵√2*x - 4 = 0

D = (2*⁵√2)² + 4*4*(2*⁵√2²) = 4*⁵√2² + 32*⁵√2² = 36*⁵√2²

x12 = (2*⁵√2 +- √(36*⁵√2²))/4*⁵√2² =  (2*⁵√2 +- 6*⁵√2))/4*⁵√2² =

- ⁵√2/⁵√2²  и 2* ⁵√2/⁵√2²  

x1 =  - ⁵√2/⁵√2²  < 0 нет

x2 = 2* ⁵√2/⁵√2²

⁵√y² =  2* ⁵√2/⁵√2²

y² = 2⁵*2/2² = 2⁴

y12 = ± √2⁴ = ± 4

y = {-4,4}