Решите уравнения (1/5)^x^2+x-10=625; (4/3)^x^2+2^2x+1=136

Решение
1)  (1/5)^(x^2+x-10) = 625
5^(- x²- x + 10) = 5⁴
- x² - x + 10 = 4
x² + x - 6 = 0
x₁ = - 2
x₂ = 3

2) 0.6^x×(5/3)^y=3/5;
(2^x)^y=64 

(3/5)^x * (3/5)^ (-y) = 3/5
(3/5)^(x - y) = 3/5
x - y = 1
y = x - 1

y = x - 1
2^x*(x - 1) = 64 

y = x - 1
2^x*(x - 1) = 2⁶
y = x - 1
x*(x - 1) = 6

y = x - 1
x² - x - 6 = 0

x² - x - 6 = 0
x₁ = - 2
x₂ = 3

x₁ = - 2
y₁ = - 3

x₂ = 3
y₂ = 2
ответ: (- 2 ; - 3)  (3 ; 2)