1. Given $a>b, a+b=2$. Then the minimum value of $\frac{a^{2}+b^{2}}{a-b}$ is ( ).
(A) $2 \sqrt{2}$
(B) $\sqrt{2}$
(C) 2
(D) 1
- 1. C.

Let $a=1+t, b=1-t(t>0)$. Then
$$
\frac{a^{2}+b^{2}}{a-b}=\frac{2\left(1+t^{2}\right)}{2 t}=t+\frac{1}{t} \geqslant 2 \text {. }
$$

Equality holds if and only if $t=1$, i.e., $a=2, b=0$.