A=\(\frac{2017^{2017}+2}{2017^{2017}-1}\)=\(\frac{\left(2017^{2017}-1\right)+3}{2017^{2017}-1}\)=\(1\)+\(\frac{3}{2017^{2017}-1}\)
B=\(\frac{2017^{2017}}{2017^{2017}-3}\)=\(\frac{\left(2017^{2017}-3\right)+3}{2017^{2017}-3}\)=\(1\)+\(\frac{3}{2017^{2017}-3}\)
Vì \(2017^{2017}-1\)\(>\)\(2017^{2017}-3\)nên \(\frac{3}{2017^{2017}-1}\)\(< \)\(\frac{3}{2017^{2017}-3}\)=> A<B
vậy A<B
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