K
Khách

Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.

6 tháng 8 2017

Ta có : \(\dfrac{2017+2018}{2018+2019}=\dfrac{2017}{2018+2019}+\dfrac{2018}{2018+2019}\)

Rõ ràng ta thấy : \(\dfrac{2017}{2018}>\dfrac{2017}{2018+2019}\) (1)

\(\dfrac{2018}{2019}>\dfrac{2018}{2018+2019}\) (2)

Từ (1)(2), suy ra :

\(\dfrac{2017}{2018}+\dfrac{2018}{2019}>\dfrac{2017+2018}{2018+2019}\)

Vậy ......................

~ Học tốt ~

6 tháng 8 2017

Ta có : \(\dfrac{2017}{2018}+\dfrac{2018}{2019}+\dfrac{2019}{2020}=\left(1-\dfrac{1}{2018}\right)+\left(1-\dfrac{1}{2019}\right)+\left(1-\dfrac{1}{2020}\right)\)\(=\left(1+1+1\right)-\left(\dfrac{1}{2018}+\dfrac{1}{2019}+\dfrac{1}{2020}\right)\)

\(=3+\left(\dfrac{1}{2018}+\dfrac{1}{2019}+\dfrac{1}{2020}\right)< 3\)

Vậy \(\dfrac{2017}{2018}+\dfrac{2018}{2019}+\dfrac{2019}{2020}< 3\)

8 tháng 9 2018

Ta có: \(B=\dfrac{2017+2018+2019}{2018+2019+2020}=\dfrac{2017}{2018+2019+2020}+\dfrac{2018}{2018+2019+2020}+\dfrac{2019}{2018+2019+2020}\)

\(\dfrac{2017}{2018}>\dfrac{2017}{2018+2019+2020}\)

\(\dfrac{2018}{2019}>\dfrac{2018}{2018+2019+2020}\)

\(\dfrac{2019}{2020}>\dfrac{2019}{2018+2019+2020}\)

\(\Rightarrow\dfrac{2017}{2018}+\dfrac{2018}{2019}+\dfrac{2019}{2020}>\dfrac{2017}{2018+2019+2020}+\dfrac{2018}{2018+2019+2020}+\dfrac{2019}{2018+2919+2020}\)

\(\Rightarrow A>B.\)

Vậy \(A>B.\)

8 tháng 8 2017

Ta có :

\(\dfrac{2017}{2018}+\dfrac{2018}{2019}+\dfrac{2019}{2020}=\left(1-\dfrac{1}{2018}\right)+\left(1-\dfrac{1}{2019}\right)+\left(1-\dfrac{1}{2020}\right)\)

\(=\left(1+1+1\right)-\left(\dfrac{1}{2018}+\dfrac{1}{2019}+\dfrac{1}{2020}\right)\)

\(=3-\left(\dfrac{1}{2018}+\dfrac{1}{2019}+\dfrac{1}{2020}\right)< 3\)

\(\Leftrightarrow\dfrac{2017}{2018}+\dfrac{2018}{2019}+\dfrac{2019}{2020}< 3\)

8 tháng 8 2017

Ta có :

\(\dfrac{2017+2018}{2018+2019}=\dfrac{2017}{2018+2019}+\dfrac{2018}{2018+2019}\)

Ta thấy :

\(\dfrac{2017}{2018}>\dfrac{2017}{2018+2019}\left(1\right)\)

\(\dfrac{2018}{2019}>\dfrac{2018}{2018+2019}\left(2\right)\)

từ \(\left(1\right)+\left(2\right)\Leftrightarrow\dfrac{2017}{2018}+\dfrac{2018}{2019}>\dfrac{2017+2018}{2018+2019}\)

14 tháng 4 2019

Đặt \(\frac{2016}{2017}\)+\(\frac{2017}{2018}\)+\(\frac{2018}{2019}\)+\(\frac{2019}{2016}\) là A

A=1-\(\frac{1}{2017}\)+1-\(\frac{1}{2018}\)+1-\(\frac{1}{2019}\)+1+\(\frac{3}{2016}\)

A=4-(\(\frac{1}{2017}\)+\(\frac{1}{2018}\)+\(\frac{1}{2019}\)-\(\frac{3}{2016}\)) Do \(\frac{1}{2017}\)+\(\frac{1}{2018}\)+\(\frac{1}{2019}\)-\(\frac{3}{2016}\)<0 =>A>4
19 tháng 5 2018

Giải:

Ta có:

\(P=\dfrac{2016}{2017}+\dfrac{2017}{2018}+\dfrac{2018}{2019}\)

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

\(\left\{{}\begin{matrix}\dfrac{2016}{2017}=\dfrac{2016}{2017}\\\dfrac{2017}{2018}=\dfrac{2017}{2018}\\\dfrac{2018}{2019}=\dfrac{2018}{2019}\end{matrix}\right.\)

\(\Leftrightarrow\dfrac{2016}{2017}+\dfrac{2017}{2018}+\dfrac{2018}{2019}=\dfrac{2016}{2017}+\dfrac{2017}{2018}+\dfrac{2018}{2019}\)

Hay \(P=Q\)

Vậy ...

22 tháng 5 2018

bạn lm sai r

\(A>\dfrac{2^{2018}}{2^{2018}+3^{2019}+5^{2020}}+\dfrac{3^{2019}}{2^{2018}+3^{2019}+5^{2020}}+\dfrac{5^{2020}}{5^{2020}+2^{2018}+3^{2019}}=1\)

\(B< \dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{2019\cdot2020}\)

\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{2019}-\dfrac{1}{2020}\)

=>B<1

=>A>B

AH
Akai Haruma
Giáo viên
12 tháng 2 2023

Lời giải:
$A=1-\frac{1}{2019}+1-\frac{1}{2020}+1-\frac{1}{2021}+1+\frac{3}{2018}$

$=4+(\frac{1}{2018}-\frac{1}{2019}+\frac{1}{2018}-\frac{1}{2020}+\frac{1}{2018}-\frac{1}{2021})$

$> 4+0+0+0+0=4$

30 tháng 8 2017

Ta có: \(\dfrac{2017}{2018}>\dfrac{2017}{2018+2019}\)

\(\dfrac{2018}{2019}>\dfrac{2018}{2018+2019}\)

=> \(\dfrac{2017}{2018}+\dfrac{2018}{2019}>\dfrac{2017+2018}{2018+2019}\)

=> A > B

30 tháng 8 2017

Ta có :

\(B=\dfrac{2017+2018}{2018+2019}=\dfrac{2017}{2018+2019}+\dfrac{2018}{2018+2019}\)

Ta thấy :

\(\dfrac{2017}{2018}>\dfrac{2017}{2018+2019}\left(1\right)\)

\(\dfrac{2018}{2019}>\dfrac{2018}{2018+2019}\left(2\right)\)

Từ \(\left(1\right)+\left(2\right)\Leftrightarrow A>B\)

1 tháng 5 2017

\(B=\dfrac{2017^{2018}-2}{2017^{2019}-2}< 1\)

Ta có :

\(B=\dfrac{2017^{2018}-2}{2017^{2019}-2}< \dfrac{2017^{2018}-2+2019}{2017^{2019}-2+2019}=\dfrac{2017^{2018}+2017}{2017^{2019}+2017}=\dfrac{2017\left(2017^{2017}+1\right)}{2017\left(2017^{2018}+1\right)}=\dfrac{2017^{2017}+1}{2017^{2018}+1}=A\)

Vậy B < A