1、
(1)f(-x)=-f(x)
-2ax+1/x^2=2ax+1/x^2
4ax=0
a=0
所以,函数的解析式:f(x)=1/x^2
(2)
令x2>x1且x1、x2属于(0,1)
则,f(x2)-f(x1)
=2a(x2-x1)+1/x2^2-1/x1^2
=2a(x2-x1)+(x1^2-x2^2)/(x1x2)^2
=(x2-x1)(2a-x1-x2)/(x1x2)^2
据假设,f(x2)-f(x1)
1、
(1)f(-x)=-f(x)
-2ax+1/x^2=2ax+1/x^2
4ax=0
a=0
所以,函数的解析式:f(x)=1/x^2
(2)
令x2>x1且x1、x2属于(0,1)
则,f(x2)-f(x1)
=2a(x2-x1)+1/x2^2-1/x1^2
=2a(x2-x1)+(x1^2-x2^2)/(x1x2)^2
=(x2-x1)(2a-x1-x2)/(x1x2)^2
据假设,f(x2)-f(x1)