通过考虑 $\left(x + \frac{1}{ax^2}\right)^7$ 二项式展开式中的通项（其中 $a$ 为常数），解释为什么每一项都与 $x$ 相关。 在 $\left(x + \frac{1}{ax^2}\right)^7$ 的展开式中，$x$ 的系数为 $\frac{7}{3}$。求 $a$ 的所有可能值。

1. Write down the general term of the binomial expansion

The general term is given by:
$T_{r+1} = \binom{7}{r} x^{7-r} \left(\frac{1}{ax^2}\right)^r$

2. Simplify the general term to isolate the power of $x$

$T_{r+1} = \binom{7}{r} x^{7-r} \left(a^{-1} x^{-2}\right)^r$
$T_{r+1} = \binom{7}{r} a^{-r} x^{7-r-2r}$
$T_{r+1} = \binom{7}{r} a^{-r} x^{7-3r}$

3. Explain why no term is independent of $x$

For a term to be independent of $x$, the power of $x$ must be $0$:
$7 - 3r = 0$
$3r = 7$
$r = \frac{7}{3}$
Since $r$ must be an integer ($0 \leqslant r \leqslant 7$), $r = \frac{7}{3}$ is not possible. Thus, there is no term independent of $x$, which means every term is dependent on $x$.

1. Find the value of $r$ that gives the term in $x$

The general term is $T_{r+1} = \binom{7}{r} x^{7-r} \left(\frac{1}{ax^2}\right)^r = \binom{7}{r} \frac{1}{a^r} x^{7-3r}$.
For the term in $x$, the power of $x$ must be 1:
$7 - 3r = 1$
$3r = 6 \implies r = 2$

2. Equate the coefficient to $\frac{7}{3}$ and solve for $a$

Substitute $r = 2$ into the coefficient part:
$\text{Coefficient} = \binom{7}{2} \frac{1}{a^2} = 21 \left(\frac{1}{a^2}\right)$
Given that the coefficient is $\frac{7}{3}$:
$\frac{21}{a^2} = \frac{7}{3}$
$a^2 = \frac{21 \times 3}{7}$
$a^2 = 9$
$a = \pm 3$