7. If real numbers $x, y$ satisfy
$$
\frac{x}{2^{10}+5^{3}}+\frac{y}{2^{10}+6^{3}}=1, \frac{x}{3^{10}+5^{3}}+\frac{y}{3^{10}+6^{3}}=1 \text {. }
$$

then $x+y=$
$$
\text { II. } 7.2^{10}+3^{10}+5^{3}+6^{3} \text {. }
$$

According to the conditions, $2^{10}$ and $3^{10}$ are the two roots of the equation with respect to $t$:
$$
\frac{x}{t+5^{3}}+\frac{y}{t+6^{3}}=1
$$

which means they are the two roots of the equation:
$$
t^{2}-\left(x+y-5^{3}-6^{3}\right) t+30^{3}-6^{3} x-5^{3} y=0
$$

Therefore,
$$
\begin{array}{l}
2^{10}+3^{10}=x+y-5^{3}-6^{3} \\
\Rightarrow x+y=2^{10}+3^{10}+5^{3}+6^{3} .
\end{array}
$$