關(guān)于x的方程f（x）-a|x-1|=0恰有4個(gè)互異的實(shí)根,
a=|(x^2+3x)/(x-1)|有4個(gè)原像,①
設(shè)u=(x^2+3x)/(x-1)=x+4+4/(x-1)=x-1+4/(x-1)+5,
|x-1+4/(x-1)|=|x-1|+4/|x-1|>=4,
∴x-1+4/(x-1)>=4,或x-1+4/(x-1)=9或u9或a=9或0