因为当x=0时f(0)=c×0/(2×0+3)=0f[f(0)]=f(0)=0所以此时与c的值无关正常的解法：f(f(x)) =c[cx/(2x+3)]/{2[cx/(2x+3)]+3} 上下乘2x+3 =c^2x/[2cx+3(2x+3)] =c^2x/[(2c+6)x+9] =x 所以c^2=(2c+6)x+9 (2c+6)x=c^2-9 ...