One, (20 points) Given real numbers $x, y$ satisfy
$$
x+y=3, \frac{1}{x+y^{2}}+\frac{1}{x^{2}+y}=\frac{1}{2} \text {. }
$$

Find the value of $x^{5}+y^{5}$.
Given $\frac{1}{x+y^{2}}+\frac{1}{x^{2}+y}=\frac{1}{2}$, we have
$$
2\left(x+y+x^{2}+y^{2}\right)=x^{3}+y^{3}+x^{2} y^{2}+x y \text {. }
$$

Let $x y=t$. Then
$$
\begin{array}{l}
x^{2}+y^{2}=(x+y)^{2}-2 x y=9-2 t, \\
x^{3}+y^{3}=(x+y)\left((x+y)^{2}-3 x y\right)=3(9-3 t) .
\end{array}
$$

Substituting into equation (1) gives
$$
2(3+9-2 t)=3(9-3 t)+t^{2}+t
$$
$\Rightarrow t=1$ or 3.
When $t=3$, $x y=3$, but $x+y=32 \sqrt{x y}$ fits the problem, at this time,
$$
\begin{array}{l}
x^{5}+y^{5}=\left(x^{2}+y^{2}\right)\left(x^{3}+y^{3}\right)-(x+y) x^{2} y^{2} \\
=(9-2 t) \times 3(9-3 t)-3 t^{2}=123 .
\end{array}
$$