Question

Найдите корни уравнения(пожалуйста)
(x^2+3)(x-7)=0
(3y-1)(y^2+1)=0
(z-1)^2(z+4)=0
(3t+12)(t+2)^2=0

Answer 1

(x^2 + 3) ( x - 7) = 0
x^2 = - 3 ==> нет реш
x - 7 = 0 ==> x = 7 

(3y - 1) (y^2  + 1) = 0
3y - 1 = 0 ==> 3y = 1 ==> y = 1/3
y^2 = - 1 ==> нет реш

(z - 1)^2 (z + 4) = 0 
z - 1 = 0 ==> z = 1
z + 4 = 0 ==> z = - 4

(3t + 12) (t + 2)^2 = 0 
3t + 12 = 0 ==> 3t = - 12  ==> t = - 4;
t + 2 = 0 ==> t = - 2