4x(x-5)=0 (3x-6)x2,4=0 (X+3)x(x+4)=0

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вопрос

The given equations are:

1) 4x(x-5) = 0
This equation will be true if either 4x = 0 or (x-5) = 0.
So, the solutions are:
4x = 0 => x = 0
x-5 = 0 => x = 5

Therefore, the solutions are x=0 and x=5.

2) (3x-6)x^2 = 0
This equation will be true if either (3x-6) = 0 or x^2 = 0.
So, the solutions are:
3x-6 = 0 => 3x = 6 => x = 2
x^2 = 0 => x = 0

Therefore, the solutions are x=0 and x=2.

3) (x+3)x(x+4) = 0
This equation will be true if either (x+3) = 0 or x = 0 or (x+4) = 0.
So, the solutions are:
x+3 = 0 => x = -3
x = 0
x+4 = 0 => x = -4

Therefore, the solutions are x=-3, x=0, and x=-4.