7. Let $a, b>0$, satisfy the equation about $x$
$$
\sqrt{|x|}+\sqrt{|x+a|}=b
$$

has exactly three distinct real solutions $x_{1}, x_{2}, x_{3}$, and $x_{1}<x_{2}<x_{3}$ $=b$. Then the value of $a+b$ is $\qquad$
7. 144.

Let $t=x+\frac{a}{2}$. Then the equation in terms of $t$ is
$$
\sqrt{\left|t-\frac{a}{2}\right|}+\sqrt{\left|t+\frac{a}{2}\right|}=b
$$

has exactly three distinct real solutions
$$
t_{i}=x_{i}+\frac{a}{2}(i=1,2,3) .
$$

Since $f(t)=\sqrt{\left|t-\frac{a}{2}\right|}+\sqrt{\left|t+\frac{a}{2}\right|}$ is an even function, the three real solutions of the equation $f(t)=b$ are symmetrically distributed about the origin on the number line. Therefore,
$$
b=f(0)=\sqrt{2 a} .
$$

Next, we solve the equation $f(t)=\sqrt{2 a}$ for real solutions.
When $|t| \leqslant \frac{a}{2}$,
$$
\begin{array}{l}
f(t)=\sqrt{\frac{a}{2}-t}+\sqrt{\frac{a}{2}+t} \\
=\sqrt{a+\sqrt{a^{2}-4 t^{2}}} \leqslant 2 a,
\end{array}
$$

with equality holding if and only if $t=0$;
When $t>\frac{a}{2}$, $f(t)$ is monotonically increasing, and when $t=\frac{5 a}{8}$, $f(t)=\sqrt{2 a}$;

When $t<-\frac{a}{2}$, $f(t)$ is monotonically decreasing, and when $t=-\frac{5 a}{8}$, $f(t)=\sqrt{2 a}$.
Thus, the equation $f(t)=\sqrt{2 a}$ has exactly three real solutions
$$
t_{1}=-\frac{5}{8} a, t_{2}=0, t_{3}=\frac{5}{8} a \text {. }
$$

By the given condition, we know $b=x_{3}=t_{3}-\frac{a}{2}=\frac{a}{8}$, combining with $b=\sqrt{2 a}$, we get $a=128$.
Therefore, $a+b=\frac{9 a}{8}=144$.