5^x+5^y=30
x+y=3
система

x = 3 - y5^(3 - y) + 5^y = 30 *125/5^y + 5^y - 30 = 0 Пусть 5^y = t125/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 = 25y₁ = 25^y = 5y₂ = 1x₁ = 1y₁ = 2x₂ = 2y₂ = 1Ответ( 1; 2) ; (2; 1)