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

To find the set of all real values of $x$ for which the given inequality holds, let us write the given expression:

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

We can solve this inequality by considering two cases based on the definition of the absolute value function $|x+9|$.

Case 1: When $x+9\ge 0$ (i.e., $x\ge -9$)

In this region, $|x+9|=x+9$.

Substituting this into the inequality, we get:

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

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

$x^2-9>0$

Factoring the left-hand side gives:

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

The solution to this quadratic inequality is:

$x<-3\text{ or }x>3$

Taking the intersection with the case condition $x\ge -9$:

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

Case 2: When $x+9<0$ (i.e., $x<-9$)

In this region, $|x+9|=-(x+9)$.

Substituting this into the inequality, we get:

$x^2-[-(x+9\left)\right]+x>0$

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

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

Let us check the discriminant $D$ of the quadratic expression $x^2+2x+9$:

$D=b^2-4ac=2^2-4\left(1\right)\left(9\right)=4-36=-32<0$

Since the coefficient of $x^2$ is positive ($1>0$) and the discriminant is negative, the quadratic expression $x^2+2x+9$ is strictly positive for all real values of $x$.

Thus, all values of $x$ in this case satisfy the inequality.

Taking the intersection with the case condition $x<-9$:

$x\in (-\infty ,-9)$

Combining the solutions from both cases:

Taking the union of the solutions from Case 1 and Case 2:

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

Combining $(-\infty ,-9)$ and $[-9,-3)$ gives:

$x\in (-\infty ,-3)\cup (3,\infty )$

Therefore, the set of all real values of $x$ satisfying the inequality is $(-\infty ,-3)\cup (3,\infty )$.