Ta có \(B=\frac{2015+2016+2017}{2016+2017+2018}\)

\(\Leftrightarrow B=\frac{2015}{2016+2017+2018}+\frac{2016}{2016+2017+2018}+\frac{2017}{2016+2017+2018}\)

Vì
\(\frac{2015}{2016}>\frac{2015}{2016+2017+2018};\frac{2016}{2017}>\frac{2016}{2016+2017+2018};\frac{2017}{2018}>\frac{2017}{2016+2017+2018}\) nên \(\frac{2015}{2016}+\frac{2016}{2017}+\frac{2017}{2018}>\frac{2015}{2016+2017+2018}+\frac{2016}{2016+2017+2018}+\frac{2017}{2016+2017+2018}\)

Hay \(A>B\)

tớ chịu thôi SORRY CẬU RẤT NHIỀU

A=2015/2016+2016/2017+2017/2018>2015/2018+2016/2018+2017/2018

=6048/2018>1

B=2015+2016+2017/2016+2017+2018=6048/6051<1

=>A>B

Có: B = 2015 + 2016 + 2017/2016 + 2017 + 2018

       B= 2015 / (2015 + 2016+2017)  + 2016/(2016+2017+2018) + 2017/(2016 + 2017 + 2018)

vì     2015/2016 > 2015/(2016 + 2017+2018)  ; 2016/2017>2016/(2016+2017+2018) ; 2017/2018 > 2017/(2016+2017+2018)

=> A>B

có ai là ARMY ko nếu là ARMY thì mọi người cày view chưa

A<B(2015/2016<2015;2016/2017<2016;2017/2018<2017)

Đặt 2015.2016+2016=n

suy ra A=(n+1)/n   và B=(n+2)/(n+1)

Ta có A - B=(n+1)/n -(n+2)/(n+1)=((n+1)²-n(n+2))/n(n+1)=(n²+2n+1-n²-2n)/n(n+1)=1/n(n+1)

Vì A-B lớn hơn 0 nên A>B

thanks