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

The correct option is:
$(-\infty ,-3)\cup (3,\infty )$

To solve the given inequality, we must eliminate the absolute value. The inequality is:

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

The expression inside the absolute value is $x+9$. Since the absolute value function behaves differently depending on whether its argument is positive or negative, we must split the problem into two distinct cases: when $x+9\ge 0$ and when $x+9<0$.

Case 1: $x+9\ge 0$

This condition implies that:
$x\ge -9$
Since $x+9$ is non-negative, the absolute value is simply the expression itself: $|x+9|=x+9$. Substituting this back into our original inequality gives:

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

Now, distribute the negative sign and simplify:

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

Taking the square root of both sides, we get:
$x<-3$  or  $x>3$

However, we must also satisfy our Case 1 condition, $x\ge -9$. By intersecting these requirements, the valid interval for Case 1 is:
$x\in [-9,-3)\cup (3,\infty )$

Case 2: $x+9<0$

This condition implies that:
$x<-9$
Since $x+9$ is negative in this case, its absolute value is its negation: $|x+9|=-(x+9)$. Substitute this back into the original inequality:

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

This simplifies to:

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

Let's analyze the quadratic equation $x^2+2x+9$ by checking its discriminant ($D=b^2-4ac$):
$D=2^2-4\left(1\right)\left(9\right)=4-36=-32$

Because the discriminant is negative ($D<0$) and the leading coefficient ($x^2$) is positive, the quadratic expression $x^2+2x+9$ is strictly greater than $0$ for all real numbers $x$.
Thus, the inequality holds true for any value in Case 2. The valid interval here is strictly determined by our case condition:
$x\in (-\infty ,-9)$

Combining the Cases

The total solution is the union of the valid intervals from Case 1 and Case 2:
$x\in (-\infty ,-9)\cup [-9,-3)\cup (3,\infty )$

Notice how the boundary $-9$ perfectly joins the first two sets, simplifying the expression to:

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