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

We are asked to solve

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

Split into two cases based on the absolute value.

Case 1: $$x+9 \ge 0 \Rightarrow x \ge -9$$

$$|x+9| = x+9$$

Inequality becomes: $$x^2 - (x+9) + x > 0 \implies x^2 - 9 > 0 \implies (x-3)(x+3) > 0$$

So $$x<-3  or  x>3$$. Combined with $$x\ge -9$$, we get $$x\in [-9,-3) \cup (3,\infty)$$

Case 2: $$x+9 < 0 \Rightarrow x < -9$$

$$|x+9| = -(x+9) = -x - 9$$

Inequality becomes: $$x^2 - (-x-9) + x > 0 \implies x^2 + 2x + 9 > 0$$

The quadratic $$x^2 + 2x + 9$$ has discriminant (4-36=-32 <0), so always positive. But in this case (x<-9), so inequality is satisfied. Thus (x < -9) also works.

$$x < -3  \text{or}  x>3$$

So the solution set is $${(-\infty,-3) \cup (3,\infty)}$$