Áp dụng bđt Svacxo ta có :

\(\dfrac{2017}{\sqrt{2018}}+\dfrac{2018}{\sqrt{2017}}\ge\dfrac{\left(\sqrt{2017}+\sqrt{2018}\right)^2}{\sqrt{2018}+\sqrt{2017}}=\sqrt{2017}+\sqrt{2018}\)

Dấu bằng xảy ra khi:

\(\dfrac{2017}{\sqrt{2018}}=\dfrac{2018}{\sqrt{2017}}\Leftrightarrow2017=2018\left(vl\right)\)

Suy ra không xảy ra dấu bằng

Vậy \(\dfrac{2017}{\sqrt{2018}}+\dfrac{2018}{\sqrt{2017}}>\sqrt{2017}+\sqrt{2018}\)

\(\frac{2017}{\sqrt{2018}}+\frac{2018}{\sqrt{2017}}=\frac{2017\sqrt{2017}+2018\sqrt{2018}}{\sqrt{2017}\cdot\sqrt{2018}}\)

\(=\left(\sqrt{2017}+\sqrt{2018}\right)\cdot\frac{2017+2018-\sqrt{2018\cdot2017}}{\sqrt{2017\cdot2018}}\)

Ta thấy \(\frac{2017+2018-\sqrt{2018\cdot2017}}{\sqrt{2018\cdot2017}}=\frac{\sqrt{2017}}{\sqrt{2018}}+\frac{\sqrt{2018}}{\sqrt{2017}}-1\)

Áp dụng ĐBT Cô si thì \(\frac{\sqrt{2017}}{\sqrt{2018}}+\frac{\sqrt{2018}}{\sqrt{2017}}\ge2\Rightarrow\frac{\sqrt{2017}}{\sqrt{2018}}+\frac{\sqrt{2018}}{\sqrt{2017}}-1\ge1\)

\(\Rightarrow\sqrt{2017}+\sqrt{2018} < \frac{2017}{\sqrt{2018}}+\frac{2018}{\sqrt{2017}}\)

a, x=\(\frac{1\left(\sqrt{2019}+\sqrt{2018}\right)}{2019-2018}\) và y=\(\frac{1\left(\sqrt{2018}+\sqrt{2017}\right)}{2018-2017}\) (Trục căn thức ở mẫu)

\(\Leftrightarrow\) x=\(\sqrt{2019}+\sqrt{2018}\) và y=\(\sqrt{2018}+\sqrt{2017}\)

b, Ta có : x - y = (\(\sqrt{2019}+\sqrt{2018}\) ) - ( \(\sqrt{2018}+\sqrt{2017}\) )

= \(\sqrt{2019}-\sqrt{2017}\) > 0

\(\Rightarrow\) x - y > 0 \(\Leftrightarrow\) x > y

Câu b đề sai nha, bây giờ đặt \(a=\sqrt{2017},b=\sqrt{2018}\)

Ta có ^{\(\frac{a^2}{b}+\frac{b^2}{a}< a+b\Leftrightarrow ab\left(\frac{a^2}{b}+\frac{b^2}{a}\right)< ab\left(a+b\right)\)}

\(\Leftrightarrow a^3+b^3< ab\left(a+b\right)\)(1)

Mà \(ab\left(a+b\right)\le\left(a^2-ab+b^2\right)\left(a+b\right)=a^3+b^3\)(2)

Từ (1), (2) => Sai

a) Ta có:

\(\frac{1}{\left(k+1\right)\sqrt{k}}=\frac{k+1-k}{\left(k+1\right)\sqrt{k}}=\frac{\left(\sqrt{k+1}+\sqrt{k}\right)\left(\sqrt{k+1}-\sqrt{k}\right)}{\left(k+1\right)\sqrt{k}}\)\(< \frac{2\sqrt{k+1}\left(\sqrt{k+1}-\sqrt{k}\right)}{\left(k+1\right)\sqrt{k}}=\frac{2\left(\sqrt{k+1}-\sqrt{k}\right)}{\sqrt{k+1}\sqrt{k}}=\frac{2}{\sqrt{k}}-\frac{2}{\sqrt{k+1}}\)

Cho k=1,2,....,n rồi cộng từng vế ta có:

\(\frac{1}{2}+\frac{1}{3\sqrt{2}}+\frac{1}{4\sqrt{3}}+....+\frac{1}{\left(n+1\right)\sqrt{n}}< \left(\frac{2}{\sqrt{1}}-\frac{2}{\sqrt{2}}\right)+\left(\frac{2}{\sqrt{2}}-\frac{2}{\sqrt{3}}\right)\)\(+\left(\frac{2}{\sqrt{3}}-\frac{2}{\sqrt{4}}\right)+....+\left(\frac{2}{\sqrt{n}}-\frac{2}{\sqrt{n+1}}\right)=2-\frac{2}{\sqrt{n-1}}< 2\)

\(A=\frac{\sqrt{2017}^2}{\sqrt{2018}}+\frac{\sqrt{2018}^2}{\sqrt{2017}}\ge\frac{\left(\sqrt{2017}+\sqrt{2018}\right)^2}{\sqrt{2018}+\sqrt{2017}}=\sqrt{2017}+\sqrt{2018}\)

Dấu "=" ko xảy ra nên \(\frac{2017}{\sqrt{2018}}+\frac{2018}{\sqrt{2017}}>\sqrt{2018}+\sqrt{2017}\)

a, \(\frac{\sqrt{2}\left(1-\sqrt{3}\right)}{1-\sqrt{3}}\)-\(\frac{3\left(1+\sqrt{3}\right)}{1+\sqrt{3}}\)

=\(\sqrt{2}-3\)

b,X=\(\sqrt{2019}+\sqrt{2018}\)

(Khử mẫu,nhân tử&mẫu vs\(\sqrt{2019}+\sqrt{2018}\))

Y=\(\sqrt{2018}+\sqrt{2017}\)

(Khử mẫu,nhân tử&mẫu vs\(\sqrt{2018}+\sqrt{2017}\))

So sánh:X & Y<=>X-\(\sqrt{2018}\)&Y-\(\sqrt{2018}\)(Trừ hai vế cho \(\sqrt{2018}\)) <=>\(\sqrt{2019}\)&\(\sqrt{2017}\)

Có:2019>2017

=>\(\sqrt{2019}>\sqrt{2017}\)

=>X>Y

Câu b, mk ko bt có lm đúng ko?