18. (本题10分)
(1) 计算:$\frac{a^{2}}{a - 1}+\frac{1}{1 - a}$.
(2) 先化简,再求值:$(\frac{2m + n}{m^{2}-mn}+\frac{1}{m})\cdot(m^{2}-n^{2})$,已知$m + n = 1$.

(1)
$\frac{a^{2}}{a - 1}+\frac{1}{1 - a}$
$=\frac{a^{2}}{a - 1}-\frac{1}{a - 1}$
$=\frac{a^{2}-1}{a - 1}$
$=\frac{(a + 1)(a - 1)}{a - 1}$
$=a + 1$
(2)
$(\frac{2m + n}{m^{2}-mn}+\frac{1}{m})\cdot(m^{2}-n^{2})$
$=[\frac{2m + n}{m(m - n)}+\frac{1}{m}]\cdot(m + n)(m - n)$
$=(\frac{2m + n}{m(m - n)}+\frac{m - n}{m(m - n)})\cdot(m + n)(m - n)$
$=\frac{2m + n + m - n}{m(m - n)}\cdot(m + n)(m - n)$
$=\frac{3m}{m(m - n)}\cdot(m + n)(m - n)$
$=3(m + n)$
当$m + n = 1$时，原式$ = 3×1=3$