化简二次根式!(a-b)√-(a-b)分之1
问题描述：
化简二次根式!(a-b)√-(a-b)分之1
答
要使√-(a-b)分之一有意义,则-1/(a-b)>0,
解得a-b所以│a-b│=-(a-b)
所以 (a-b)√-(a-b)分之1
= (a-b)√[-(a-b)/(a-b)^2]
= (a-b)*(1/│a-b│)√[-(a-b)]
= (a-b)*[-1/(a-b)]√[-(a-b)]
=-√[-(a-b)]