The set of all real values of $x$ for which $(x^2 - |x + 9| + x) > 0$, is

Since there is an absolute value $|x + 9|$, split into two cases based on the sign of $(x + 9)$.

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When $x \geq -9$, $|x+9| = x+9$:

$x^2 - (x + 9) + x > 0$

$x^2 - x - 9 + x > 0$

$x^2 - 9 > 0$

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

This is positive when $x < -3$ or $x > 3$.

Applying the restriction $x \geq -9$, the solution from this case is $[-9, -3) \cup (3, \infty)$.

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When $x < -9$, $|x+9| = -(x+9)$:

$x^2 + (x+9) + x > 0$

$x^2 + 2x + 9 > 0$

$(x+1)^2 + 8 > 0$

Since $(x+1)^2 \geq 0$, this expression is always $\geq 8 > 0$, so the inequality holds for all $x < -9$.

The solution from this case is $(-\infty, -9)$.

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Combining both cases:

$(-\infty, -9) \cup [-9, -3) \cup (3, \infty)$

$= (-\infty, -3) \cup (3, \infty)$