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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
Xét B = \(\frac{19^{31}+5}{19^{32}+5}< \frac{19^{31}+5+14}{19^{32}+5+14}=\frac{19^{31}.19}{19^{32}.19}=\frac{19\left(19^{30}+1\right)}{19\left(19^{31}+1\right)}=\frac{19^{30}+1}{19^{31}+1}< \frac{19^{30}+5}{19^{31}+5}=A\)Vậy 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
\(A=\frac{19^5-1+2017}{19^5-1}=1+\frac{2017}{19^5-1}\)
\(B=\frac{19^5+2015}{19^5-2}=\frac{19^5-2+2017}{19^5-2}=1+\frac{2017}{19^5-2}\)
\(\Rightarrow1+\frac{2017}{19^5-1}< 1+\frac{2017}{19^5-2}\)
\(\Rightarrow A< B\)
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.....