Решите уравнение: (9y+18) (12y-4) (36y-72) = 0

(9y+18) (12y-4) (36y-72) = 0

9 (y+2) 4 (3y-1) 36 (y-2) = 0

1296 (y+2) (y-2) (3y-1) = 0

(y+2) (y-2) (3y-1) = 0

первый корень:

y+2=0

y=-2

второй корень:

y-2=0

y=2

третий корень:

3y-1 = 0

3y=1

y=1/3

(9y+18) (12y-4) (36y-72) = 0

9 (y+2) 4 (3y-1) 36 (y-2) = 0

1296 (y+2) (y-2) (3y-1) = 0

(y+2) (y-2) (3y-1) = 0

y+2=0

y1=-2

y-2=0

y2=2

3y-1 = 0

3y=1

y3=1/3