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Resolver la siguiente in ecuación |2x-3|< 3+x-x^2

User Langkiller
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Answer:

Explanation:

To solve the equation, we'll need to consider two cases:

Case 1: 2x - 3 is positive or zero

If 2x - 3 is positive or zero, then |2x - 3| = 2x - 3, and the inequality becomes:

2x - 3 < 3 + x - x^2

Rearranging and simplifying:

x^2 - x - 6 < 0

Factoring:

(x - 3)(x + 2) < 0

The solutions to this inequality are:

-2 < x < 3

However, we still need to check that 2x - 3 is indeed positive or zero for this range of x. We can see that this is true for x in the range (-2, 3), so this is a valid solution.

Case 2: 2x - 3 is negative

If 2x - 3 is negative, then |2x - 3| = -(2x - 3), and the inequality becomes:

-(2x - 3) < 3 + x - x^2

Rearranging and simplifying:

x^2 - 3x - 6 < 0

Factoring:

(x - 3)(x + 2) > 0

The solutions to this inequality are:

x < -2 or x > 3

However, we still need to check that 2x - 3 is indeed negative for this range of x. We can see that this is true for x < -2, so this is a valid solution.

Putting these two cases together, we get the solution:

x < -2 or -2 < x < 3

I hope this helps! Let me know if you have any questions.

User Mameesh
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