\(A=\frac{19^5-1+2017}{19^5-1}\)

\(A=1+\frac{2017}{19^5-1}\)

\(B=\frac{19^5+2015}{19^5-2}=\frac{19^5-2+2017}{19^5-2}=\frac{1+2017}{19^5-2}\)

\(\Rightarrow1+\frac{2017}{19^5-1}< 1+\frac{2017}{19^5-2}\)

\(\Rightarrow A< B\)

LẤY (2015/19^5-1)-(2014/19^5-2)=(2015*19^5-2*2015-2014*19^5+2014)/((19^5-10*(19^5-2)

=(19^5-2016)/((19^5-1)*(19^5-2)>0

HAY A>B

phan conan ơi dấu sao là j

Nhân tích chéo 

mình chỉ gợi ý thôi bạn tự làm nhé

\(A=\frac{19^5-1+2017}{19^5-1}=1+\frac{2017}{19^5-1}\)

\(B=\frac{19^5+2015}{19^5-2}=\frac{19^5-2+2017}{19^5-2}=1+\frac{2017}{19^5-2}\)
\(\Rightarrow1+\frac{2017}{19^5-1}< 1+\frac{2017}{19^5-2}\)
\(\Rightarrow A< B\)

ta thấy:B>1

=>\(B=\frac{19^5+2015}{19^5-2}>\frac{19^5+2015+1}{19^5-2+1}=\frac{19^5+2016}{19^5-1}=A\Rightarrow B>A\)

vậy.....

Ta có: \(A=\frac{19^5+2016}{19^5-1}=\frac{19^5-1+2017}{19^5-1}=\frac{19^5-1}{19^5-1}+\frac{2017}{19^5-1}=1+\frac{2017}{19^5-1}\)

\(B=\frac{19^5+2015}{19^5-2}=\frac{19^5-2+2017}{19^5-2}=\frac{19^5-2}{19^5-2}+\frac{2017}{19^5-2}=1+\frac{2017}{19^5-2}\)

Vì \(\frac{2017}{19^5-1}< \frac{2017}{19^5-2}\Rightarrow1+\frac{2017}{19^5-1}< 1+\frac{2017}{19^5-2}\Rightarrow A< B\)

Vậy A < B

\(A=\frac{19^5+2016}{19^5-1}=\frac{\left(19^5-1\right)+2017}{19^5-1}=1+\frac{2017}{19^5-1}\)

\(B=\frac{19^5+2015}{19^5-2}=\frac{\left(19^5-2\right)+2017}{19^5-2}=1+\frac{2017}{19^5-2}\)

Vì \(19^5-1>19^5-2\) nên \(\frac{2017}{19^5-1}< \frac{2}{19^5-2}\)

\(\Rightarrow1+\frac{2017}{19^5-1}< 1+\frac{2017}{19^5-2}\)

Vậy \(A< B\)

Ta có: \(A=\dfrac{19^5+2016}{19^5-1}=1+\dfrac{2017}{19^5-1}\)

\(B=\dfrac{19^5+2015}{19^5-2}=1+\dfrac{2017}{19^5-2}\)

Vì \(\dfrac{2017}{19^5-1}< \dfrac{2017}{19^5-2}\Rightarrow1+\dfrac{2017}{19^5-1}< 1+\dfrac{2017}{19^5-2}\)

\(\Rightarrow A< B\)

Vậy A < B

A < B

\(A< B\)