已知函数f(x)=ax^2+2ax+4(a>0)若m

因为：f(m)=am^2+2am+4f(n)=an^2+2an+4所以：f(m)-f(n)=(am^2+2am+4)-(an^2+2an+4)=a(m^2-n^2)+2a(m-n)=a(m-n)(m+n)+2a(m-n)=a(m-n){(m+n)+2}=a(m-n)(a-1+2)=a(m-n)(a+1)又 由 a>0 m0 m-n