5. Let the equation $a x^{2}+(a+2) x+9 a=0$ with respect to $x$ have two distinct real roots $x_{1} 、 x_{2}$, and $x_{1}<\frac{2}{5}<x_{2}$
(A) $a>0$
(B) $-\frac{2}{7}<a<0$
(C) $a<-\frac{2}{7}$
(D) $-\frac{2}{11}<a<0$
5. (D).

It is clear that $a \neq 0$. The original equation can be transformed into
$$
x^{2}+\left(1+\frac{2}{a}\right) x+9=0 \text {. }
$$

Let $y=x^{2}+\left(1+\frac{2}{a}\right) x+9$, then this parabola opens upwards. Since $x_{1}<1<x_{2}$, when $x=1$, $y<0$. That is, $1+\left(1+\frac{2}{a}\right)+9<0$. Solving this, we get $-\frac{2}{11}<a<0$.