设曲线y=x+1/x-1在点(3,2)处的切线与直线ax+y+3=0垂直,则a=?

方法一:y=(x+1)/(x-1)=1+[2/(x-1)]y'=-2/(x-1)²当x=3时,y'=-2/(3-1)²=-1/2该点切线与直线ax+y+3=0垂直即它们的斜率乘积为-1-1/2×(-a)=-1解得a=-2,二:y=(x+1)/(x-1)=1+2/(x-1)在(3,2)的切线：记为y=k(x-3)...那我这张试卷就靠你了(#∩_∩#)请考虑答题者的辛苦，点击“满意”，谢谢