Question

(x + 2,7)(x - 4)(x + 6) = 0 х¹=-2,7 х²=4 х³= - 6 из них целые только х² и х³, поэтому А={4, -6}

Answer 1

возможно так, но это никак не точно)

Answer 2

Let's verify the solution step by step.

We are given the equation:

(x + 2)(x - 4)(x + 6) = 0

To find the solutions for x, we set each factor equal to zero:

1. (x + 2) = 0 x = -2

2. (x - 4) = 0 x = 4

3. (x + 6) = 0 x = -6

Now, we check which of these values make x^1, x^2, and x^3 integers:

1. For x = -2: x^1 = -2 (an integer) x^2 = (-2)^2 = 4 (an integer) x^3 = (-2)^3 = -8 (not an integer)

2. For x = 4: x^1 = 4 (an integer) x^2 = 4^2 = 16 (an integer) x^3 = 4^3 = 64 (an integer)

3. For x = -6: x^1 = -6 (an integer) x^2 = (-6)^2 = 36 (an integer) x^3 = (-6)^3 = -216 (not an integer)

From the above calculations, we can see that only x^2 and x^3 are integers for x = 4 and x = -6, respectively.

Therefore, the set A should be: A = {4, -6}. The initial statement is correct.