⚡ AZUCATION SHORTCUT:
Value Substitution:
If $x = -10$: $100 - |-1| - 10 = 89 > 0$ (Correct). Options C and A (which stops at -9) are out.
If $x = -5$: $25 - |4| - 5 = 16 > 0$ (Correct). Option D is out as it skips $[-9, -3]$.
Only Option B covers all valid negative values.

Case 1: $x \ge -9$
Expression: $x^2 - (x + 9) + x > 0$
$x^2 - x - 9 + x > 0 \implies x^2 - 9 > 0$
$(x-3)(x+3) > 0$. Roots are $3, -3$.
Solution for this case: $x \in [-9, -3) \cup (3, \infty)$.

Case 2: $x < -9$
Expression: $x^2 - (-(x + 9)) + x > 0$
$x^2 + x + 9 + x > 0 \implies x^2 + 2x + 9 > 0$
For $x^2 + 2x + 9$, $D = 2^2 - 4(9) = -32$. Since $D < 0$ and $a > 0$, the quadratic is always positive for all real $x$.
Solution for this case: $x \in (-\infty, -9)$.

Final Union:
Combining $(-\infty, -9)$ and $[-9, -3) \cup (3, \infty)$ gives us the complete set:
$x \in \mathbf{(-\infty, -3) \cup (3, \infty)}$.

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