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 \(x>y\)

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 \(A=\frac{2017-2018}{2017+2018}=\frac{\left(2017-2018\right)\left(2017+2018\right)}{\left(2017+2018\right)^2}=\frac{2017^2-2018^2}{2017^2+2018^2+2.2017.2018}< \frac{2017^2-2018^2}{2017^2+2018^2}=B\)

Vậy A<B

A=24783,14746B=49566,29188

Vậy A<B

Ta thấy \(A=\frac{2018-2017}{2018+2017}=\frac{2018^2-2017^2}{\left(2018+2017\right)^2}=\frac{2018^2-2017^2}{2018^2+2.2018.2017+2017^2}\)

Mà \(2018^2+2.2018.2017+2017^2>2018^2+2017^2\)

\(\Rightarrow\frac{2018^2-2017^2}{2018^2+2.2018.2017+2017^2}< \frac{2018^2-2017^2}{2018^2+2017^2}\)

Vậy 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

Đặt \(2017=a\)

\(A=\sqrt{1+a^2+\dfrac{a^2}{\left(a+1\right)^2}}+\dfrac{a}{a+1}\\ A=\sqrt{\left(a+1\right)^2-2a+\dfrac{a^2}{\left(a+1\right)^2}}+\dfrac{a}{a+1}\\ A=\sqrt{\left(a+1\right)^2-2\left(a+1\right)\cdot\dfrac{a}{a+1}+\left(\dfrac{a}{a+1}\right)^2}+\dfrac{a}{a+1}\\ A=\sqrt{\left(a+1-\dfrac{a}{a+1}\right)^2}+\dfrac{a}{a+1}\\ A=\left|a+1-\dfrac{a}{a+1}\right|+\dfrac{a}{a+1}\)

Ta có \(\dfrac{a}{a+1}< 1\Leftrightarrow a+1-\dfrac{a}{a+1}>0\)

\(\Leftrightarrow A=a+1-\dfrac{a}{a+1}+\dfrac{a}{a+1}=a+1=2018\)

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

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