$1 \cdot 34$ If real numbers $x, y$ satisfy $x^{2}+y^{2}+x^{2} y^{2}-4 x y+1=0$, then the value of $(x+y)^{2}$ is
(A) 1.
(B) 4.
(C) 9.
(D) 16.
(China Sichuan Province Junior High School Mathematics Competition, 1991)
[Solution] The given equation can be transformed into
that is, $\square$
$$
\begin{array}{c}
\left(x^{2}+y^{2}-2 x y\right)+\left(x^{2} y^{2}-2 x y+1\right)=0, \\
(x-y)^{2}+(x y-1)^{2}=0 .
\end{array}
$$

Then $x-y=0$, and $x y-1=0$, i.e., $x y=1$. Thus, $(x+y)^{2}=(x-y)^{2}+4 x y=4$. Therefore, the answer is (B).