The set of all real numbers x for which x^2 - |x + 2| + x > 0 is 1. (-?, -2) ? (2, ?) 2. (-?, -?2) ? (?2, ?)3. (-?, -1) ? (1, ?) 4. (?2, ?)"

The condition given in the question is x^2 – |x + 2| + x > 0

Two cases are possible:

Case 1: When (x+2) ≥ 0.

Therefore, x^2 – x – 2 + x > 0

Hence, x^2 – 2 > 0

So, either x √2.

Hence, x ∈ [-2, -√2) ∪ (√2, ∞) ……. (1)

Case 2: When (x+2)

Then x^2 + x + 2 + x > 0

So, x^2 + 2x + 2 > 0

This gives (x+1)^2 + 1 > 0 and this is true for every x

Hence, x ≤ -2 or x ∈ (-∞, -2) ……… (2)

From equations (2) and (3) we get x ∈ (-∞, -√2) ∪ (√2, ∞).

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