因为分母是 $|x+8|$，所以要以 $x=-8$ 为分界点分情况讨论。注意 $x=-8$ 本身不能取。

在每个区间内把绝对值去掉，再解普通二次不等式。

We need

where

If $x>-8$, then $|x+8|=x+8$. Since $x+8>0$,

Factorise:

So, in this case,

This gives

Combining with $x>-8$ gives

If $x<-8$, then $|x+8|=-(x+8)$. Since $|x+8|>0$, multiply by $|x+8|$:

Compute the right hand side:

Therefore

Factorise:

Thus

so

Combining with $x<-8$ gives

Therefore the full solution is

另一种思路是先找出所有临界点（即等式成立的边界值以及无定义点），然后在数轴上划分区间进行代值测试。此法可免去在不等式中处理负号和不等号方向变号的困扰。

等式方程为 $\dfrac{x^2-9}{|x+8|} = 6-2x$，因为绝对值恒为非负，故可直接变形为 $x^2-9 = (6-2x)|x+8|$，再依 $x > -8$ 和 $x < -8$ 分类找出所有解。

First, identify all critical values. The vertical asymptote (where the expression is undefined) is:

Now solve the equation for the boundary points of intersection:

Since $|x+8| > 0$ for all $x \ne -8$, this is equivalent to:

Case 1: $x > -8$, which means $|x+8| = x+8$:

Both values are greater than $-8$, so they are valid critical values.

Case 2: $x < -8$, which means $|x+8| = -(x+8)$:

Since we require $x < -8$, the only valid root in this case is:

Thus, the critical values are:

These values divide the real number line into five intervals. Testing a value from each interval in the inequality $\dfrac{x^2-9}{|x+8|} > 6-2x$ yields:

• For $x < -13$: e.g., $x = -14 \implies \dfrac{187}{6} \approx 31.17 > 34$ (False)
• For $-13 < x < -8$: e.g., $x = -10 \implies \dfrac{91}{2} = 45.5 > 26$ (True)
• For $-8 < x < -\dfrac{19}{3}$: e.g., $x = -7 \implies 40 > 20$ (True)
• For $-\dfrac{19}{3} < x < 3$: e.g., $x = 0 \implies -\dfrac{9}{8} > 6$ (False)
• For $x > 3$: e.g., $x = 4 \implies \dfrac{7}{12} \approx 0.58 > -2$ (True)

Therefore, the inequality holds on: