Question

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

Answer 1

(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

Answer 2

(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

Answer 3

To solve the given equation, we need to find the values of 'y' that make the expression (9y+18)(12y-4)(36y-72) equal to zero. When the product of three factors is zero, at least one of the factors must be zero.

So, we set each factor to zero and solve for 'y':

1. 9y + 18 = 0 9y = -18 y = -18 / 9 y = -2

2. 12y - 4 = 0 12y = 4 y = 4 / 12 y = 1 / 3

3. 36y - 72 = 0 36y = 72 y = 72 / 36 y = 2

The solutions to the equation (9y+18)(12y-4)(36y-72) = 0 are y = -2, y = 1/3, and y = 2.