解：先化简分式（k2-1）/(9-k2)+（2k+1）/(k+3)=(1-k2）/(k2-9)+[（2k+1）(k-3)]/(k2-9)=(k2-5k-2)/(k2-9)方程化为(9x-3)/(x^2-x)=（x+k）/(x^2-x)方程无解,有增根为x=0或x=1使9x-3=x+k当x=0k=-3原式分式无意义当x=1k=5原分式式=-1/8
