38. Given $\boldsymbol{a}=(1, x), \boldsymbol{b}=\left(x^{2}+x,-x\right), m$ is a constant and $m \leqslant-2$, find the range of $x$ that satisfies $\boldsymbol{a} \cdot \boldsymbol{b}+2>m\left(\frac{2}{\boldsymbol{a} \cdot \boldsymbol{b}}+1\right)$.
$$\begin{array}{l}
\text { 38. } \because \boldsymbol{a}=(1, x), \boldsymbol{b}=\left(x^{2}+x,-x\right), \\
\quad \therefore \boldsymbol{a} \cdot \boldsymbol{b}=x^{2}+x-x^{2}=x . \\
\text { Therefore, } \boldsymbol{a} \cdot \boldsymbol{b}+2>m\left(\frac{2}{\boldsymbol{a} \cdot \boldsymbol{b}}+1\right) \\
\quad \Leftrightarrow x+2>m\left(\frac{2}{x}+1\right) \Leftrightarrow(x+2)-m \frac{x+2}{x}>0 \\
\Leftrightarrow \frac{(x+2)(x-m)}{x}>0 \\
\quad \Leftrightarrow x(x+2)(x-m)>0 .
\end{array}$$
(1) When $m=-2$, the range of $x$ is $(0,+\infty)$; (2) When $m<-2$, the range of $x$ is $(m,-2) \cup(0,+\infty)$.