\(A=\left(\frac{x-1}{\sqrt{x}-1}+\frac{x+2\sqrt{x}+1}{\sqrt{x}+1}\right).\frac{1}{2\sqrt{x}}=\left[\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x-1}}+\frac{\left(\sqrt{x}+1\right)^2}{\sqrt{x}+1}\right].\frac{1}{2\sqrt{x}}\)

\(A=2\left(\sqrt{x}+1\right).\frac{1}{2\sqrt{x}}=\frac{\sqrt{x}+1}{\sqrt{x}}>1=\sqrt{\frac{2019}{2019}}>\sqrt{\frac{2018}{2019}}\) ( đpcm ) 

... 

ta xét : \(\sqrt{a^2+b^2+\frac{a^2}{\left(\frac{a}{b}+1\right)^2}}=\sqrt{\left(a+b\right)^2-2ab+\frac{a^2b^2}{\left(a+b\right)^2}}=\sqrt{\left(a+b\right)^2-2.\left(a+b\right).\frac{ab}{a+b}+\frac{a^2b^2}{\left(a+b\right)^2}}=\sqrt{\left(a+b-\frac{ab}{a+b}\right)^2}=\left|a+b-\frac{ab}{a+b}\right|\)

áp dụng vào bài toán :

\(A=\left|1+2018-\frac{2018}{2019}\right|+\frac{2018}{2019}=2019\)

Đặt:   \(A=1+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+....+\frac{1}{\sqrt{2018}}\)

Ta có:   \(\frac{1}{\sqrt{k}}=\frac{2}{\sqrt{k}+\sqrt{k}}>\frac{2}{\sqrt{k}+\sqrt{k+1}}=2\left(\sqrt{k+1}-\sqrt{k}\right)\)       với \(\forall k\inℕ^∗\)

Do đó ta có:    \(A>2\left[\left(\sqrt{2019}-\sqrt{2018}\right)+\left(\sqrt{2018}-\sqrt{2017}\right)+...+\left(\sqrt{3}-\sqrt{2}\right)\right]+1\)

  \(=2\left(\sqrt{2019}-\sqrt{2}\right)+1=2\sqrt{2019}-2\sqrt{2}+1>2\sqrt{2019}-3+1>2\sqrt{2019}-2\)

   \(>2\sqrt{2018}-2=2\left(\sqrt{2018}-1\right)\)

=>  đpcm

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:

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