7. Given the sequence $\left\{x_{n}\right\}$ satisfies:
$$
\begin{array}{l}
x_{1}=1, \\
x_{i+1}-x_{i}=\sqrt{x_{i+1}+x_{i}}(i=1,2, \cdots) .
\end{array}
$$

Then the general term formula $x_{n}=$
7. $x_{n}=\frac{n^{2}+n}{2}$.

From $x_{i+1}-x_{i}=\sqrt{x_{i+1}+x_{i}}$
$\Rightarrow x_{i+1}-x_{i} \geqslant 0$.
Also, $x_{1}=1$, so $x_{i+1} \geqslant x_{i} \geqslant 1(i=1,2, \cdots)$.
Squaring both sides of equation (1) and simplifying, we get
$$
x_{i+1}^{2}-\left(2 x_{i}+1\right) x_{i+1}+x_{i}^{2}-x_{i}=0 \text {. }
$$

Using the quadratic formula, we have
$$
x_{i+1}=\frac{2 x_{i}+1 \pm \sqrt{8 x_{i}+1}}{2} \text {. }
$$

If $x_{i+1}=\frac{2 x_{i}+1-\sqrt{8 x_{i}+1}}{2}$, then
$$
x_{i+1}-x_{i}=\frac{1-\sqrt{8 x_{i}+1}}{2}<0,
$$

which contradicts equation (2).
$$
\begin{array}{l}
\text { Therefore, } x_{i+1}=\frac{2 x_{i}+1+\sqrt{8 x_{i}+1}}{2} \\
\Rightarrow 8 x_{i+1}=8 x_{i}+4 \sqrt{8 x_{i}+1}+4 \\
\Rightarrow 8 x_{i+1}+1=\left(\sqrt{8 x_{i}+1}+2\right)^{2} \\
\Rightarrow \sqrt{8 x_{i+1}+1}=\sqrt{8 x_{i}+1}+2 \\
\Rightarrow \sqrt{8 x_{n}+1}=3+2(n-1) \\
\Rightarrow x_{n}=\frac{n^{2}+n}{2} .
\end{array}
$$