f(X)=x平方除以ax+b方程f(X)-x+12=0有两个实根x1=3 .x2=4设k>1解x不等式f(X)<x(k+1)-k除以2-k ?

f(x)-x+12=0
x^2/(ax+b)-x+12=0
x^2=(ax+b)(x-12),(ax+b不等于0)
因为这个方程有两个实根,则分别代入方程,得到
3a+b=-1
4a+b-2
解之得,a=-1,b=2

f(x)=x^2/(ax+b)=x^2/(-x+2)

f(x)<x(k+1)/2-x
x^2/(2-x)<(kx-x)/2

x<2时,2x^2<(2-x)(k-1)x
x((k+1)x-2(k-1))<0
0<x<2(k-1)/(k+1)

x>2时, 2x^2>(2-x)(k-1)x
x((k+1)x-2(k-1))>0
x<0,或x>2(k-1)/(k+1),且x不等于2