关于x的分式方程a/(x-1)(x+1)-1/x-1=2/x+2有曾根,则a=

两边同时乘以(x-1)(x+1)(x+2)得：a(x+2)-(x+1)(x+2)=2(x-1)(x+1)即a(x+2)=3x^2+3x增根只可能为1,-1,-2分别代入得：a*3=3+3,得：a=2a*1=3-3,得：a=0a*8=12-6,无解但a=0时,方程化为：-1/(x-1)=2/(x+2),得：x=0不为增根...