Ta có: \(\frac{1}{\left(k+1\right)\sqrt{k}+k\sqrt{k+1}}=\frac{\left(k+1\right)\sqrt{k}-k\sqrt{k+1}}{k\left(k+1\right)^2-k^2\left(k+1\right)}\)

\(=\frac{\sqrt{k\left(k+1\right)}\left(\sqrt{k+1}-\sqrt{k}\right)}{k^3+2k^2+k-k^3-k^2}\)

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

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

Lần lượt thay k=1;2;...;2018 ta được:

\(\frac{1}{2\sqrt{1}+1\sqrt{2}}=\frac{1}{1}-\frac{1}{\sqrt{2}}\)

\(\frac{1}{3\sqrt{2}+2\sqrt{3}}=\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}\)

...

\(\frac{1}{2019\sqrt{2018}+2018\sqrt{2019}}=\frac{1}{\sqrt{2018}}-\frac{1}{\sqrt{2019}}\)

Cộng vế theo vế ta được:

\(C=1-\frac{1}{\sqrt{2019}}=...\)

\(\frac{1}{\sqrt{k\left(2018-k+1\right)}}>\frac{2}{k+2019-k}=\frac{2}{2019}\)

Ap dụng bài toan được

\(A>\frac{2}{2019}+\frac{2}{2019}+...+\frac{2}{2019}=2.\frac{2018}{2019}\)

Với mọi \(n\inℕ^∗\) ta có:

 \(\frac{1}{\left(n+1\right)\sqrt{n}+n\sqrt{n+1}}=\frac{\left(n+1\right)\sqrt{n}-n\sqrt{n-1}}{\left(n+1\right)^2n-n^2\left(n+1\right)}\)

\(=\frac{\left(n+1\right)\sqrt{n}-n\sqrt{n+1}}{n\left(n+1\right)}=\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n-1}}\)

Áp dụng đẳng thức trên ta có:

\(A=\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+....+\frac{1}{\sqrt{2018}}-\frac{1}{\sqrt{2019}}\)

\(=1-\frac{1}{\sqrt{2019}}\)

   \(t\text{ổng}qu\text{át}:\frac{1}{n\sqrt{n-1}+\left(n-1\right)\sqrt{n}}=\frac{n\sqrt{n-1}-\left(n-1\right)\sqrt{n}}{n^2\left(n-1\right)-\left(n-1\right)^2n}\)

\(=\frac{n\sqrt{n-1}-\left(n-1\right)\sqrt{n}}{\left(n-1\right)n}\)

\(=\frac{1}{\sqrt{n-1}}-\frac{1}{\sqrt{n}}\)

Thay vào A có

\(A=\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-...+\frac{1}{\sqrt{2016}}-\frac{1}{\sqrt{2017}}\)

\(=1-\frac{1}{\sqrt{2017}}\)

1)

DKCĐ: a>0,\(a\ne1\)

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