Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
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) và (2), suy ra :
\(\dfrac{2017}{2018}+\dfrac{2018}{2019}>\dfrac{2017+2018}{2018+2019}\)
Vậy ......................
~ Học tốt ~
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 \(\dfrac{2017}{2018}+\dfrac{2018}{2019}+\dfrac{2019}{2020}< 3\)
Ta có :
\(\dfrac{2017}{2018}+\dfrac{2018}{2019}+\dfrac{2019}{2020}=\left(1-\dfrac{1}{2018}\right)+\left(1-\dfrac{1}{2019}\right)+\left(1-\dfrac{1}{2020}\right)\)
\(=\left(1+1+1\right)-\left(\dfrac{1}{2018}+\dfrac{1}{2019}+\dfrac{1}{2020}\right)\)
\(=3-\left(\dfrac{1}{2018}+\dfrac{1}{2019}+\dfrac{1}{2020}\right)< 3\)
\(\Leftrightarrow\dfrac{2017}{2018}+\dfrac{2018}{2019}+\dfrac{2019}{2020}< 3\)
Giải:
Ta có:
\(P=\dfrac{2016}{2017}+\dfrac{2017}{2018}+\dfrac{2018}{2019}\)
và \(Q=\dfrac{2016}{2017}+\dfrac{2017}{2018}+\dfrac{2018}{2019}\)
Vì \(\left\{{}\begin{matrix}\dfrac{2016}{2017}=\dfrac{2016}{2017}\\\dfrac{2017}{2018}=\dfrac{2017}{2018}\\\dfrac{2018}{2019}=\dfrac{2018}{2019}\end{matrix}\right.\)
\(\Leftrightarrow\dfrac{2016}{2017}+\dfrac{2017}{2018}+\dfrac{2018}{2019}=\dfrac{2016}{2017}+\dfrac{2017}{2018}+\dfrac{2018}{2019}\)
Hay \(P=Q\)
Vậy ...
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
Ta có: \(B=\dfrac{2017+2018+2019}{2018+2019+2020}=\dfrac{2017}{2018+2019+2020}+\dfrac{2018}{2018+2019+2020}+\dfrac{2019}{2018+2019+2020}\)
Mà \(\dfrac{2017}{2018}>\dfrac{2017}{2018+2019+2020}\)
\(\dfrac{2018}{2019}>\dfrac{2018}{2018+2019+2020}\)
\(\dfrac{2019}{2020}>\dfrac{2019}{2018+2019+2020}\)
\(\Rightarrow\dfrac{2017}{2018}+\dfrac{2018}{2019}+\dfrac{2019}{2020}>\dfrac{2017}{2018+2019+2020}+\dfrac{2018}{2018+2019+2020}+\dfrac{2019}{2018+2919+2020}\)
\(\Rightarrow A>B.\)
Vậy \(A>B.\)
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
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 \(\left(1\right)+\left(2\right)\Leftrightarrow A>B\)
Vì \(B=\dfrac{2017^{2018}-2}{2017^{2019}-2}< 1\)
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
Lời giải:
$A=1-\frac{1}{2019}+1-\frac{1}{2020}+1-\frac{1}{2021}+1+\frac{3}{2018}$
$=4+(\frac{1}{2018}-\frac{1}{2019}+\frac{1}{2018}-\frac{1}{2020}+\frac{1}{2018}-\frac{1}{2021})$
$> 4+0+0+0+0=4$
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 \(\left(1\right)+\left(2\right)\Leftrightarrow\dfrac{2017}{2018}+\dfrac{2018}{2019}>\dfrac{2017+2018}{2018+2019}\)