5^x+5^y=30

x+y=3

система

X = 3 - y

5^ (3 - y) + 5^y = 30 *

125/5^y + 5^y - 30 = 0

Пусть 5^y = t

125/t + t - 30 = 0

t^2 - 30t + 125 = 0

D = 900 - 500 = 400

t₁ = (30 + 20) / 2 = 25;

t₂ = (30 - 20) / 2 = 5;

5^y = 25

y₁ = 2

5^y = 5

y₂ = 1

x₁ = 1

y₁ = 2

x₂ = 2

y₂ = 1

Ответ

(1; 2) ; (2; 1)