$解:(1)x_{1}=c,x_{2}=\frac{m}{c}(m≠0)$
$(2)結(jié)論:方程x+\frac {n}{x}=c+\frac {n}{c}(n≠0)的$
$解為x_{1}=c,x_{2}=\frac{n}{c}(n≠0)$
$關(guān) 于x的方程化為即x-1+\frac{2}{x-1}=a-1+\frac{2}{a-1}$
$則x-1=a-1或x-1=\frac{2}{a-1}$
$解得x_{1}=a,x_{2}=\frac {2}{a-1}+1=\frac{a+1}{a-1}$