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 \(A>B\)

Bạn Linh lẽ ra phải chứng minh như vầy đã chứ A=2015/2016  +  2016/2017=( 1 - 1/2016) + ( 1 - 1/2017)= 2 - 1/2016 - 1/2017 > 1

doán bua a>b

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

Ta có 

1 - A = 1 - 2014/2015 = 1/2015

1 - B = 1 - 2015/ 2016 = 1/2016 

Vì  1/2015 > 1/2016 => 1 - 2014/2015 > 1 - 2015 / 2016 

Hay 1 - A > 1 -B =>   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

ta có 2015/2016+2016/2017+2017/2015=(1-1/2016)+(1-1/2017)+(2+1/2015)

                                                         =4-(1/2016+1/2017-1/2015)

                   1/2016<1; 1/2017<1 nên 1/2016+1/2017<2 suy ra 1/2016+1/2017-1/2015<1(vì 1/2015<1)

4-(1/2016+1/2017-1/2015)>4-1=3           

2015/2016+2016/2017+2017/2015>3

cho mik nhé