33. If it is known that $a$ and $b$ are opposite numbers, and $|a-b|=2$, then the value of $\frac{a^{2}+2 a b+2 b^{2}+2 a+2 b+1}{a^{2}+3 a b+b^{2}+3}$ is
Answer: 1
Solution: Since $a, b$ are opposite numbers, we have $a+b=0$, thus
$$
\frac{a^{2}+2 a b+2 b^{2}+2 a+2 b+1}{a^{2}+3 a b+b^{2}+3}=\frac{(a+b+1)^{2}+b^{2}}{a b+3}=\frac{1+b^{2}}{a b+3},
$$

From $|a-b|=2$, we get when $a-b=2$,

When $a-b=-2$,

In summary, we get
In summary, we get
$$
\begin{array}{l}
a-b= \pm 2 \\
a=1, b=-1,
\end{array}
$$
$$
\begin{array}{c}
a b=-1, b^{2}=1, \\
\frac{1+b^{2}}{a b+3}=1 ;
\end{array}
$$
$$
\begin{array}{c}
a=-1, b=1, \\
a b=-1, b^{2}=1, \\
\frac{1+b^{2}}{a b+3}=1 \\
\frac{a^{2}+2 a b+2 b^{2}+2 a+2 b+1}{a^{2}+3 a b+b^{2}+3}=1 .
\end{array}
$$