Cho \(M=\dfrac{2018^{2017}+1}{2018^{2018}+1}\) và \(N=\dfrac{2018^{2016}+1}{2018^{2017}+1}\)
So sánh M và N
Giúp mk nha now!
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Những câu hỏi liên quan
18 tháng 4 2017
a, Ta có: \(\dfrac{2016}{2017+2018}< \dfrac{2016}{2017}\)
\(\dfrac{2017}{2017+2018}< \dfrac{2017}{2018}\)
\(\Rightarrow A=\dfrac{2016+2017}{2017+2018}< B=\dfrac{2016}{2017}+\dfrac{2017}{2018}\)
Vậy A < B
b, Ta có: \(\dfrac{2017}{2016+2017}< \dfrac{2017}{2016}\)
\(\dfrac{2018}{2016+2017}< \dfrac{2018}{2017}\)
\(\Rightarrow M=\dfrac{2017+2018}{2016+2017}< N=\dfrac{2017}{2016}+\dfrac{2018}{2017}\)
Vậy M < N
NK
1
T
8 tháng 6 2019
#)Giải :
\(Q=2+\frac{2016}{2017+2018+2019}+\frac{2017}{2017+2018+2019}+\frac{2018}{2017+2018+2019}\)
Ta thấy : \(2>\frac{2016}{2017};2>\frac{2017}{2018};2>\frac{2018}{2019}\left(1\right)\)
\(\frac{2016}{2017+2018+2019}< \frac{2016}{2017}\left(2\right)\)
\(\frac{2017}{2017+2018+2019}< \frac{2017}{2018}\left(3\right)\)
\(\frac{2018}{2017+2018+2019}< \frac{2018}{2019}\left(4\right)\)
Từ (1) (2) (3) (4) \(\Rightarrow P>Q\)
8 tháng 10 2021
Áp dụng BĐT Cauchy–Schwarz ta được:
\(x=\dfrac{2017}{\sqrt{2018}}+\dfrac{2018}{\sqrt{2017}}\ge\dfrac{\left(\sqrt{2018}+\sqrt{2017}\right)^2}{\sqrt{2018}+\sqrt{2017}}=\sqrt{2018}+\sqrt{2017}=y\)
Dấu \("="\Leftrightarrow\dfrac{2017}{\sqrt{2018}}=\dfrac{2018}{\sqrt{2017}}\Leftrightarrow2017=2018\left(vô.lí\right)\)
Vậy đẳng thức ko xảy ra hay \(x>y\)
AK
10 tháng 4 2018
Ta có :
\(\frac{2016}{2017}>\frac{2016}{2017+2018+2019}\)
\(\frac{2017}{2018}>\frac{2017}{2017+2018+2019}\)
\(\frac{2018}{2019}>\frac{2018}{2017+2018+2019}\)
\(\Rightarrow\frac{2016}{2017}+\frac{2017}{2018}+\frac{2018}{2019}>\) \(\frac{2016}{2017+2018+2019}+\frac{2017}{2017+2018+2019}+\frac{2018}{2017+2018+2019}\)
\(\Rightarrow P>\frac{2016+2017+2018}{2017+2018+2019}\)
\(\Rightarrow P>Q\)
Chúc bạn học tốt !!!
10 tháng 4 2018
vì P có các số bé hơn 1 còn Q có các số lớn hơn 1 =>P<Q
Vậy P<Q.
mình làm hơi tắt xin bạn thông cảm bạn tự viết các số có trong P;Q ra nhá
18 tháng 3 2018
\(A=\dfrac{\dfrac{1}{2017}+\dfrac{2}{2016}+\dfrac{3}{2015}+...+\dfrac{2016}{2}+\dfrac{2017}{1}}{\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2016}+\dfrac{1}{2017}+\dfrac{1}{2018}}\)
\(A=\dfrac{\left(\dfrac{1}{2017}+1\right)+\left(\dfrac{2}{2016}+1\right)+\left(\dfrac{3}{2015}+1\right)+...+\left(\dfrac{2016}{2}+1\right)+1}{\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2016}+\dfrac{1}{2017}+\dfrac{1}{2018}}\)
\(A=\dfrac{\dfrac{2018}{2017}+\dfrac{2018}{2016}+\dfrac{2018}{2015}+...+\dfrac{2018}{2}+\dfrac{2018}{2018}}{\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2016}+\dfrac{1}{2017}+\dfrac{1}{2018}}\)
\(A=\dfrac{2018\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2016}+\dfrac{1}{2017}+\dfrac{1}{2018}\right)}{\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2016}+\dfrac{1}{2017}+\dfrac{1}{2018}}=2018\)
Ta có :
\(M=\dfrac{2018^{2017}+1}{2018^{2018}+1}< 1\)
\(\Rightarrow M< \dfrac{2018^{2017}+1+2017}{2017^{2018}+1+2017}=\dfrac{2018^{2017}+2018}{2017^{2018}+2018}=\dfrac{2018\left(2018^{2016}+1\right)}{2018\left(2018^{2017}+1\right)}=\dfrac{2018^{2016}+1}{2018^{2017}+1}=N\)
\(\Rightarrow M< N\)
Giải:
Ta có:
\(2018M=\dfrac{\left(2018^{2017}+1\right)2018}{2018^{2018}+1}.\)
\(2018M=\dfrac{2018^{2018}+2018}{2018^{2018}+1}.\)
\(2018M=\dfrac{\left(2018^{2018}+1\right)+2017}{2018^{2018}+1}.\)
\(2018M=\dfrac{2018^{2018}+1}{2018^{2018}+1}+\dfrac{2017}{2018^{2018}+1}.\)
\(2018M=1+\dfrac{2017}{2018^{2018}+1}._{\left(1\right)}\)
Ta lại có:
\(2018N=\dfrac{\left(2018^{2016}+1\right)2018}{2018^{2017}+1}.\)
\(2018N=\dfrac{2018^{2017}+2018}{2018^{2017}+1}.\)
\(2018N=\dfrac{\left(2018^{2017}+1\right)+2017}{2018^{2017}+1}.\)
\(2018N=\dfrac{2018^{2017}+1}{2018^{2017}+1}+\dfrac{2017}{2018^{2017}+1}.\)
\(2018N=1+\dfrac{2017}{2018^{2017}+1}._{\left(2\right)}\)
Và \(\dfrac{2017}{2018^{2018}+1}< \dfrac{2017}{2018^{2017}+1}._{\left(3\right)}\)
Từ \(_{\left(1\right);\left(2\right)}\) và \(_{\left(3\right)}\Rightarrow2018M< 2018N\Rightarrow M< N.\)
Vậy......
~ Học tốt!!! ~