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

Ta có:2015/2016>2015/2016+2017+2018

2016/2017>2016/2016+2017+2018

2017/2018>2017/2016+2017+2018-Mình áp dụng so sánh phân số cùng tử đấy.

Suy ra2015/2016+2016/2017+2017/2018>(2015+2016+2017)/(2016+2017+2018)=B

#)Giải :

\(Q=2+\frac{2016}{2017+2018+2019}+\frac{2017}{2017+2018+2019}+\frac{2018}{2017+2018+2019}\)

Ta thấy : \(2>\frac{2016}{2017};2>\frac{2017}{2018};2>\frac{2018}{2019}\left(1\right)\)

\(\frac{2016}{2017+2018+2019}< \frac{2016}{2017}\left(2\right)\)

\(\frac{2017}{2017+2018+2019}< \frac{2017}{2018}\left(3\right)\)

\(\frac{2018}{2017+2018+2019}< \frac{2018}{2019}\left(4\right)\)

Từ (1) (2) (3) (4) \(\Rightarrow P>Q\)

Xét: \(\frac{\left(17^{2017}+16^{2017}\right)^{2018}}{17^{2017.2018}}=\left(\frac{17^{2017}+16^{2017}}{17^{2017}}\right)^{2018}=\left(1+\left(\frac{16}{17}\right)^{2017}\right)^{2018}\)

\(\frac{\left(17^{2018}+16^{2018}\right)^{2017}}{17^{2017.2018}}=\left(\frac{17^{2018}+16^{2018}}{17^{2018}}\right)^{2017}=\left(1+\left(\frac{16}{17}\right)^{2018}\right)^{2017}\)

Ta có: \(0< \frac{16}{17}< 1\)

=> \(\left(\frac{16}{17}\right)^{2017}>\left(\frac{16}{17}\right)^{2018}\)

=> \(1+\left(\frac{16}{17}\right)^{2017}>1+\left(\frac{16}{17}\right)^{2018}>1\)

=> \(\left(1+\left(\frac{16}{17}\right)^{2017}\right)^{2018}>\left(1+\left(\frac{16}{17}\right)^{2018}\right)^{2017}\)

=> \(\left(17^{2017}+16^{2017}\right)^{2018}>\left(17^{2018}+16^{2018}\right)^{2017}\)

a, Bn quy đồng rồi làm nha

b,Có  A=2017^2017+1/2017^2018+1

--> 2017A=2017^2018+2017/2017^2018+1

2017A=2017^2018+1/2017^2018+1 + 2016/2017^2018+1

2017A=1+ 2016/2017^2018+1

Có B=2017^2016+1/2017^2017+1

--> 2017B=2017^2017+2017/2017^2017+1

2017B=2017^2017+1/2017^2017+1 + 2016/2017^2017+1

2017B=1+2016/2017^2017+1

        Vì 1+2016/2017^2018+1 < 1+2016/2017^2017+1

nên 2017A<2017B

-->A<B

A.Ta có : 

\(A=-\frac{15}{46}>-\frac{15}{45}=-\frac{51}{153}>-\frac{51}{151}=B\)

\(\Rightarrow A>B\)

Câu b, giống bạn Kelly Gaming TV 

B = \(\frac{2015+2016+2017}{2016+2017+2018}=\frac{2016.3}{2017.3}=\frac{2016}{2017}\left(1\right)\)

Mà A = \(\frac{2015}{2016}+\frac{2016}{2017}+\frac{2017}{2018}.\left(2\right)\)

Từ \(\left(1\right)\)và \(\left(2\right)\)=> A > B.

Vậy A > B . 

Bạn Dont look at me

Bạn nên làm theo bạn ấy

Bạn k đúng cho bạn ấy. Bởi vì bạn ấy làm đúng

Theo mk là vậy

\(A=\frac{2015}{2016}+\frac{2016}{2017}+\frac{2017}{2018}\)\(B=\frac{2015+2016+2017}{6051}\)

\(A=\frac{2015}{2016}+\frac{2016}{2017}+\frac{2017}{2018}\)\(B=\frac{2015}{6051}+\frac{2016}{6051}+\frac{2017}{6051}\)

=> A > B

\(A=\frac{10^{2016}+2018}{10^{2017}+2018}\)

\(\Rightarrow10A=\frac{10^{2017}+20180}{10^{2017}+2018}\)

\(=\frac{10^{2017}+2018+18162}{10^{2017}+2018}\)

\(=\frac{10^{2017}+2018}{10^{2017}+2018}+\frac{18162}{10^{2017}+2018}\)

\(=1+\frac{18162}{10^{2017}+2018}\)

\(B=\frac{10^{2017}+2018}{10^{2018}+2018}\)

\(\Rightarrow10B=\frac{10^{2018}+20180}{10^{2018}+2018}\)

\(=\frac{10^{2018}+2018+18162}{10^{2018}+2018}\)

\(=\frac{10^{2018}+2018}{10^{2018}+2018}+\frac{18162}{10^{2018}+2018}\)

\(=1+\frac{18162}{10^{2018}+2018}\)

Ta thấy: \(1+\frac{18162}{10^{2017}+2018}>1+\frac{18162}{10^{2018}+2018}\)

=> 10A > 10B

=> A > B