7. In the Cartesian coordinate system, the parabola
$$
y=x^{2}+m x-\frac{3}{4} m^{2}(m>0)
$$

intersects the $x$-axis at points $A$ and $B$. If the distances from points $A$ and $B$ to the origin are $O A$ and $O B$, respectively, and satisfy $\frac{1}{O B}-\frac{1}{O A}=\frac{2}{3}$, then the value of $m$ is $\qquad$.
7. 2 .

Let the roots of the equation $x^{2}+m x-\frac{3}{4} m^{2}=0$ be $x_{1}$ and $x_{2}$, and $x_{1}<0<x_{2}$.
From $\frac{1}{O B}-\frac{1}{O A}=\frac{2}{3}$, we know $O A>O B$.
Also, since $m>0$, the axis of symmetry of the parabola is to the left of the $y$-axis, thus,
$$
O A=\left|x_{1}\right|=-x_{1}, O B=x_{2} .
$$

Therefore, $\frac{1}{x_{2}}+\frac{1}{x_{1}}=\frac{2}{3}$,
$$
\text { then } \frac{x_{1}+x_{2}}{x_{1} x_{2}}=\frac{2}{3}=\frac{-m}{-\frac{3}{4} m^{2}} \text {. }
$$

Solving for $m$ yields $m=2$.