1.9 Simplify
$$
\frac{b x\left(a^{2} x^{2}+2 a^{2} y^{2}+b^{2} y^{2}\right)}{b x+a y}+\frac{a y\left(a^{2} x^{2}+2 b^{2} x^{2}+b^{2} y^{2}\right)}{b x+a y}
$$

The algebraic value is
(A) $a^{2} x^{2}+b^{2} y^{2}$.
(B) $(a x+b y)^{2}$.
(C) $(a x+b y)(b x+a y)$.
(D) $2\left(a^{2} x^{2}+b^{2} y^{2}\right)$.
(E) $(b x+a y)^{2}$.
(39th American High School Mathematics Examination, 1988)
$$
\text { [Solution] } \begin{aligned}
& \frac{b x\left(a^{2} x^{2}+2 a^{2} y^{2}+b^{2} y^{2}\right)}{b x+a y}+\frac{a y\left(a^{2} x^{2}+2 b^{2} x^{2}+b^{2} y^{2}\right)}{b x+a y} \\
= & \frac{(b x+a y)\left(a^{2} x^{2}+b^{2} y^{2}\right)+b x \cdot 2 a^{2} y^{2}+a y \cdot 2 b^{2} x^{2}}{b x+a y} \\
= & \frac{(b x+a y)\left(a^{2} x^{2}+b^{2} y^{2}\right)+2 a b x y(a y+b x)}{b x+a y} \\
= & a^{2} x^{2}+b^{2} y^{2}+2 a b x y . \\
= & (a x+b y)^{2} .
\end{aligned}
$$

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