y=2x/(3x-1)
=>
x=y/(3y-2)
=>
f-1(x)=x/(3x-2)
=>
f-1[f(x)]
=(2x/(3x-1))/(3*(2x/(3x-1))-2)
=x
x不为1/3
f[f-1(x)]
=2*(x/(3x-2))/(3*(x/(3x-2))-1)
=x
x不为2/3
虽然两个函数表达式相同,但定义域不同,所以不同
__________________好就给点分++==3Q
2460188273
y=2x/(3x-1)
=>
x=y/(3y-2)
=>
f-1(x)=x/(3x-2)
=>
f-1[f(x)]
=(2x/(3x-1))/(3*(2x/(3x-1))-2)
=x
x不为1/3
f[f-1(x)]
=2*(x/(3x-2))/(3*(x/(3x-2))-1)
=x
x不为2/3
虽然两个函数表达式相同,但定义域不同,所以不同
__________________好就给点分++==3Q
2460188273