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

\(A=1-\frac{1}{2016}+1-\frac{1}{2017}+1-\frac{1}{2018}+1+\frac{3}{2015}\)

\(A=4-\left(\frac{1}{2016}+\frac{1}{2017}+\frac{1}{2018}-\frac{3}{2015}\right)\)

Xét : 

\(\frac{1}{2016}< \frac{1}{2015}\)\(;\)\(\frac{1}{2017}< \frac{1}{2015}\)\(;\)\(\frac{1}{2018}< \frac{1}{2015}\)

\(\Rightarrow\)\(\frac{1}{2016}+\frac{1}{2017}+\frac{1}{2018}< \frac{1}{2015}+\frac{1}{2015}+\frac{1}{2015}\)

\(\Leftrightarrow\)\(\frac{1}{2016}+\frac{1}{2017}+\frac{1}{2018}-\frac{3}{2015}< 0\)

Suy ra : \(A=4-\left(\frac{1}{2016}+\frac{1}{2017}+\frac{1}{2018}-\frac{3}{2015}\right)>4-0=4\) ( đpcm ) 

... 

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

\(=\left(1-\frac{1}{2016}\right)+\left(1-\frac{1}{2017}\right)+\left(1-\frac{1}{2018}\right)+\left(1+\frac{3}{2015}\right)\)

\(=1-\frac{1}{2016}+1-\frac{1}{2017}+1-\frac{1}{2018}+1+\frac{1}{2015}+\frac{1}{2015}+\frac{1}{2015}\)

\(=\left(1+1+1+1\right)+\left(\left(\frac{1}{2015}-\frac{1}{2016}\right)+\left(\frac{1}{2015}-\frac{1}{2017}\right)+\left(\frac{1}{2015}-\frac{1}{2018}\right)\right)\)

\(=4+\left(\frac{1}{2015}-\frac{1}{2016}\right)+\left(\frac{1}{2015}-\frac{1}{2017}\right)+\left(\frac{1}{2015}-\frac{1}{2018}\right)\)

Vì \(\frac{1}{2015}>\frac{1}{2016};\frac{1}{2015}>\frac{1}{2017};\frac{1}{2015}>\frac{1}{2018}\)

\(\Rightarrow\frac{1}{2015}-\frac{1}{2016}>0;\frac{1}{2015}-\frac{1}{2017}>0;\frac{1}{2015}-\frac{1}{2018}>0\)

\(\Rightarrow\left(\frac{1}{2015}-\frac{1}{2016}\right)+\left(\frac{1}{2015}-\frac{1}{2017}\right)+\left(\frac{1}{2015}-\frac{1}{2018}\right)>0\)

\(\Rightarrow4+\left(\frac{1}{2015}-\frac{1}{2016}\right)+\left(\frac{1}{2015}-\frac{1}{2017}\right)+\left(\frac{1}{2015}-\frac{1}{2018}\right)>4\)

\(\Rightarrow A>4\left(dpcm\right)\)

Vì:

khi tính bài toán 2015/2016 + 2016/2017 + 2017/2018 + 2018/2019 + 2019/2020 + 2020/2015 này ra thì ta được con số là 6,000003688 con số này phải lớn hơn số 6 nên:  6,000003688 > 6

Vì:khi tính bài toán 2015/2016+2016/2017+2017/2018+2018/2019+ 2019/2020+2020/2015 ta ra được là: 6,000003688 nên: 6,000003688 > 6

2015/2016 + 2016/2017 + 2017/2018 + 2018/2019 + 2019/2020 + 2020/2015 = 6,000003688 vậy: 6,000003688 > 6

Sai đề bạn ơi, A>4 không thể xảy ra

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

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