f(x)=cx/2x+3(x≠-3/2）满足f[f(x)]=x 则常数c等于

f(x)=cx/2x+3(x≠-3/2）满足f[f(x)]=x 则常数c等于
f(x)=cx/2x+3(x≠-3/2）满足f[f(x)]=x 则常数c等于

f(x)=cx/2x+3(x≠-3/2）满足f[f(x)]=x 则常数c等于
设y=f(x)
则y=f(x)=cx/(2x+3)
y=cx/(2x+3)
x=f[f(x)]=f(y)=cy/(2y+3)
所以
cx=2xy+3y
cy=2xy+3x
两式相减得：
c(x-y)=3(y-x)
所以c=-3
如果不懂,请Hi我,祝学习愉快!