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

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 ...

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 \(x>y\)

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+\(\frac{3}{2016}\)

\(A=\dfrac{\dfrac{1}{2017}+\dfrac{2}{2016}+\dfrac{3}{2015}+...+\dfrac{2016}{2}+\dfrac{2017}{1}}{\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2016}+\dfrac{1}{2017}+\dfrac{1}{2018}}\)

\(A=\dfrac{\left(\dfrac{1}{2017}+1\right)+\left(\dfrac{2}{2016}+1\right)+\left(\dfrac{3}{2015}+1\right)+...+\left(\dfrac{2016}{2}+1\right)+1}{\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2016}+\dfrac{1}{2017}+\dfrac{1}{2018}}\)

\(A=\dfrac{\dfrac{2018}{2017}+\dfrac{2018}{2016}+\dfrac{2018}{2015}+...+\dfrac{2018}{2}+\dfrac{2018}{2018}}{\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2016}+\dfrac{1}{2017}+\dfrac{1}{2018}}\)

\(A=\dfrac{2018\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2016}+\dfrac{1}{2017}+\dfrac{1}{2018}\right)}{\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2016}+\dfrac{1}{2017}+\dfrac{1}{2018}}=2018\)

Ta có : 

\(\frac{2016}{2017}>\frac{2016}{2017+2018+2019}\)

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

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

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

\(\Rightarrow P>\frac{2016+2017+2018}{2017+2018+2019}\)

\(\Rightarrow P>Q\)

Chúc bạn học tốt !!! 

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

Đơn giản P < Q

Vì Nhìn sơ qua ta thấy tổng P gồm các phân số bé hơn 1

Tổng Q có 3 phân số lớn hơn 1

k mk đi mà làm ơnnnnnnnnnn

Tính A và B rồi ta đi so sánh:

A = \(\frac{2016}{2017}\) + \(\frac{2017}{2018}\) = \(1.999008674\)

B = \(\frac{2016+2017}{2017+2018}\) = \(0.9995043371\)

Mà 1.999008674 > 0.9995043371

Nên: A > B

Giải như bạn Trần Nhật Quỳnh thà không làm còn hơn.

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

Vì \(\hept{\begin{cases}\frac{2015}{2016}>\frac{2015}{2016+2017+2018}\\\frac{2016}{2017}>\frac{2016}{2016+2017+2018}\\\frac{2017}{2018}>\frac{2017}{2016+2017+2018}\end{cases}}\)

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

\(\Rightarrow P>Q\)

Vậy P > Q

Explosion !