1. \(\frac{2016}{2017}\)+\(\frac{2017}{2018}\)>1

2. A>B

Ta có:\(\frac{2016}{2017}< 1\)

\(\frac{2017}{2018}< 1\)

\(\frac{2018}{2019}< 1\)

\(\Rightarrow\frac{2016}{2017}+\frac{2017}{2018}+\frac{2018}{2019}>1+1+1=3\)

Vậy ......

Tham khảo nha \(https://www.olm.vn/hoi-dap/question/1216047.html\)

\(\frac{2015}{2016}+\frac{2016}{2017}+\frac{2017}{2015}\)

\(=\left(1-\frac{1}{2016}\right)+\left(1-\frac{1}{2017}\right)+\left(1+\frac{2}{2015}\right)\)

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

\(=3-\left(\frac{1}{2016}+\frac{1}{2017}\right)+\frac{2}{2015}\)

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

=> \(\frac{1}{2016}+\frac{1}{2017}< \frac{2}{2015}\)

=> \(-\left(\frac{1}{2016}+\frac{1}{2017}\right)+\frac{2}{2015}>0\)

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

trước tiên ta rút gọn 2 phân số 2015/2016+2016/2017

            TA RÚT GỌN 2016 LẠI VỚI NHAU = 2015/1 +1/2017

sau đó ta rút gọn 2 phân số 1/2017 + 2017/2015

           TA RÚT GỌN 2017 LẠI VỚI NHAU = 1/1 + 1/2015

TA CÓ: 2015/1 + 1/1 + 1/2015=2015/1 + 1 + 1/015=1/1 + 1 + 1/1= 1+1+1 = 3(VÌ TA RÚT GỌN 2015 LẠI VỚI NHAU)

VÌ: 3 = 3 

Vậy:2015/2016 + 2016/2017 + 2017/2015 = 3

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

\(B=1-\frac{1}{2017}+1-\frac{1}{2018}+1+\frac{2}{2016}\)

\(B=\left(1+1+1\right)-\left(\frac{1}{2017}+\frac{1}{2018}-\frac{2}{2016}\right)\)

\(B=3-\left(...\right)< 3\)

P/s :

\(\left(...\right)la`\left(\frac{1}{2017}+\frac{1}{2018}-\frac{2}{2016}\right)\)

quên ^^

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

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

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

Ta có:

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

\(\frac{2016}{2017}>\frac{2016}{2016+2017+2018}\)

\(\frac{2017}{2018}>\frac{2017}{2016+2017+2018}\)

Cộng vế theo vế, ta có:

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

\(hay\frac{2015}{2016}+\frac{2016}{2017}+\frac{2017}{2018}>\frac{2015+2016+2017}{2016+2017+2018}\)

\(\Rightarrow A>B\)

Vậy A >  B

N = \(\frac{2016+2017}{2017+2018}=\frac{2016}{2017+2018}+\frac{2017}{2017+2018}\)

Ta có: \(\frac{2016}{2017}>\frac{2016}{2017+2018}\)

\(\frac{2017}{2016}>\frac{2017}{2017+2018}\)

Nên M > N

Ta thấy : \(\frac{2016+2017}{2017+2018}\)=\(\frac{2016}{2017+2018}\)+\(\frac{2017}{2017+2018}\)

Vì : \(\frac{2016}{2017}\)>\(\frac{2016}{2017+2018}\)

\(\frac{2017}{2018}\)>\(\frac{2017}{2017+2018}\)

Cộng vế với vế ta được : \(\frac{2016}{2017}\)+\(\frac{2017}{2018}\)> \(\frac{2016}{2017+2018}\)+\(\frac{2017}{2017+2018}\)

Hay M > N

Vậy M > N

Chúc bạn hok tốt !!

=.....nha các bn. k mình nha

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

Mà \(\frac{2015}{2016}>\frac{2015}{2016+2017+2018}\)

       \(\frac{2016}{2017}>\frac{2016}{2016+2017+2018}\)

        \(\frac{2017}{2018}>\frac{2017}{2016+2017+2016}\)

Cộng vế theo vế, ta có : 

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

\(\Rightarrow A>B\)

Vì phân số A\(=\frac{2016^{2017}+1}{2017^{2018}+1}< 1\) mà B\(=\frac{2017^{2018}+1}{2017^{2017}+1}>1\)

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

Vậy A<B

a<1<b

=>A<b