2. For non-empty sets $A$ and $B$, define the operation:
$A \vee B=\{x \mid x \in A \cup B$, and $x \notin A \cap B\}$.
Given two intervals $M=(a, b)$ and $P=(c, d)$, where $a+b<c+d$, $ab=cd<0$. Then $M \vee P=$ $\qquad$
2. $(a, c) \cup(b, d)$.

Let $a b=c d=t<0$, then $a<0<b, c<0<d$.
$$
\begin{array}{l}
\text { Let } f(x)=x^{2}-(a+b) x+t, \\
g(x)=x^{2}-(c+d) x+t .
\end{array}
$$

Thus, $a$ and $b$ are the roots of the equation $f(x)=0$, and $c$ and $d$ are the roots of the equation $g(x)=0$.
$$
\begin{array}{l}
\text { Also, } f(c)=c^{2}-(a+b) c+t \\
=c^{2}-(a+b) c+c d \\
=c[(c+d)-(a+b)]<0,
\end{array}
$$

then $a<c<b$.
Similarly, $c<b<d$.
Therefore, $a<c<b<d$.
Hence, $M \vee P=(a, c) \cup(b, d)$.