1. Given real numbers $a, b$ satisfy $\frac{1}{a}+\frac{1}{b}=1, M=$ $\frac{b}{a}+\frac{a}{b}, N=\frac{b}{a^{2}}+\frac{a}{b^{2}}$. Then the relationship between $M$ and $N$ is ( ).
(A) $M>N$
(B) $M=N$
(C) $M<N$
(D) Uncertain
,$- 1 . \mathrm{A}$.
Since $\frac{1}{a}+\frac{1}{b}=1$, therefore, $a+b=a b$. Hence
$$
\begin{array}{l}
M=\frac{a^{2}+b^{2}}{a b}, \\
N=\frac{a^{3}+b^{3}}{a^{2} b^{2}}=\frac{(a+b)\left(a^{2}-a b+b^{2}\right)}{a^{2} b^{2}} \\
=\frac{a^{2}-a b+b^{2}}{a b}=\frac{a^{2}+b^{2}}{a b}-1=M-1 .
\end{array}
$$

Therefore, $M>N$.