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
此式當(dāng)x≠-3/2時恒成立
所以2c+6=c^2-9=0
所以c=-3