7. Let $P(x)=x^{4}+a x^{3}+b x^{2}+c x+d$, where
$a, b, c, d$ are real coefficients. Assume
$$
P(1)=7, P(2)=52, P(3)=97 \text {, }
$$

then $\frac{P(9)+P(-5)}{4}=$ $\qquad$ . (Vietnam)
7. 1202.

Notice that, $52-7=97-52=45$,
$$
\begin{array}{l}
7=45 \times 1-38,52=45 \times 2-38, \\
97=45 \times 3-38 .
\end{array}
$$

Let $Q(x)=P(x)-45 x+38$. Then $Q(x)$ is a quartic polynomial with leading coefficient 1, and
$$
Q(1)=Q(2)=Q(3)=0 \text {. }
$$

Thus, for some $r$,
$$
\begin{array}{l}
Q(x)=(x-1)(x-2)(x-3)(x-r) . \\
\text { Therefore, } \frac{1}{4}(P(9)+P(-5)) \\
=\frac{1}{4}(Q(9)+Q(-5))+26 \\
=\frac{1}{4}(8 \times 7 \times 6(9-r)+6 \times 7 \times 8(5+r))+26 \\
=\frac{1}{4}(6 \times 7 \times 8 \times 14)+26=1202 .
\end{array}
$$