5. The solution set of the inequality $\left|1+x+\frac{x^{2}}{2}\right|<1$ is
A. $\{x \mid-1<x<1\}$
B. $\left\{x \left\lvert\,-\frac{3}{2}<x<0\right.\right\}$
C. $\left\{x \left\lvert\,-\frac{5}{4}<x<0\right.\right\}$
D. $\{x \mid-2<x<0\}$
5. $\mathrm{D}$ Hint: The original inequality is equivalent to $\left(x+\frac{x^{2}}{2}\right)\left(x+\frac{x^{2}}{2}+2\right)<0$, which is $x(x+2)\left[(x+1)^{2}+3\right]<0$.