log3x+log3(x+2) = 1
ОДЗ: x> 0 и x+2 > 0
x>0 и x> -2 => x > 0
log3[x*(x+2)] = 1
log3[x^2+2x] = 1
x^2+2x = 3
x^2+2x-3=0
D = b^2 - 4ac
D = 2^2-4*1*(-3) = 4 + 12 = 16 = 4^2
x_1 = (-2 + sqrt(16))/2 = (-2+4)/2 = 2/2=1 > 0
x_2 = (-2 - sqrt(16))/2 = (-2-4)/2 = -6/2=-3 < 0 =>
=> x_2 = -3 - не корень исходного уравнения
Ответ: x = 1