Ta có : A= \(\dfrac{1}{1^m}\) +\(\dfrac{1}{1^n}\)

Và B=\(\dfrac{2015-2}{2015^m}+\dfrac{2015+2}{2015^n}\)

\(\Rightarrow\)\(\dfrac{1}{m}-\dfrac{2}{2015^m}+\dfrac{1}{n}+\dfrac{2}{2015^n}\)

\(\Rightarrow\dfrac{1}{n}+\dfrac{1}{m}+\dfrac{2\left(n-m\right)}{2015^{mn}}\)

TH1 2(n-m) >0 \(\Rightarrow\) 2015^mn >0 \(\Rightarrow\) A>B

TH2 2(n-m)<0\(\Rightarrow\) 2015^mn<0\(\Rightarrow\) A<B

TH3 2(n-m)=0\(\Rightarrow\) 2015^mn=0 \(\Rightarrow\) A=B

Xong rồi nấm ơi, bảo uyên nữa nhé