1.97 If $P=\sqrt{1988 \cdot 1989 \cdot 1990 \cdot 1991+1}+\left(-1989^{2}\right)$, then the value of $P$ is
(A) 1987.
(B) 1988.
(C) 1989.
(D) 1990.
（1st “Hope Cup” National Mathematics Invitational Competition, 1990）
[Solution] Let $x=1989$, then the expression under the radical can be written as
$$
\begin{array}{l}
(x-1) x(x+1)(x+2)+1 \\
=\left(x^{2}+x\right)\left(x^{2}+x-2\right)+1 \\
= {\left[\left(x^{2}+x-1\right)+1\right]\left[\left(x^{2}+x-1\right)-1\right]+1 } \\
=\left(x^{2}+x-1\right)^{2}-1^{2}+1 \\
=\left(x^{2}+x-1\right)^{2}, \\
\therefore \quad P=\sqrt{\left(1989^{2}+1989-1\right)^{2}}+\left(-1989^{2}\right)=1988 .
\end{array}
$$

Therefore, the correct choice is $(B)$.