M~1+1+1=3

N~1

=> M>N

m=n m>n m<n 1 trong 3 chắc chắn đúng ahihi =)))
 

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

2. A>B

Vì 2016²⁰¹⁶ + 1 < 2016²⁰¹⁷ + 1

\(\Rightarrow\frac{2016^{2016}+1}{2016^{2017}+1}< \frac{2016^{2016}+1+2015}{2016^{2016}+1+2015}=\frac{2016^{2016}+2016}{2016^{2017}+2016}=\frac{2016\left(2016^{2015}+1\right)}{2016\left(2016^{2016}+1\right)}\)

\(=\frac{2016^{2015}+1}{2016^{2016}+1}=B\)

\(\Rightarrow\)A < B

Ta có :

\(A=\frac{2016^{2016}+1}{2016^{2017}+1}< \frac{2016^{2016}+2015+1}{2016^{2017}+2015+1}=\frac{2016^{2016}+2016}{2016^{2017}+2016}=\frac{2016.\left(2016^{2015}+1\right)}{2016.\left(2016^{2016}+1\right)}\)

\(=\frac{2016^{2015}+1}{2016^{2016}+1}=B\)

\(\Rightarrow A< B\)

\(A=\frac{2016^{2016}+1}{2016^{2017}+1}< \frac{2016^{2016}+2015+1}{2016^{2017}+2015+1}\)

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

\(=\frac{2016.\left(2016^{2015}+1\right)}{2016.\left(2016^{2016}+1\right)}\)

Mả \(\frac{2016^{2015}+1}{2016^{2016}+1}=B\)

\(\Leftrightarrow A< B\)

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

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

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

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

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

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

3=3 nên tổng trên bằng 3

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

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

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

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

=.....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 : 

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

\(\Rightarrow A>B\)

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

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

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

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

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

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

⇒A>B

Chúc bạn học tốt :")

Dễ thấy B<1.

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

\(\frac{1}{2016}+\frac{1}{2017}+\frac{1}{2018}< \frac{1}{3}+\frac{1}{3}+\frac{1}{3}=1\)

Vậy A>2.

Vậy A>B.