Đặt A=\(\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{3}}+\dfrac{1}{\sqrt{4}}+...+\dfrac{1}{\sqrt{100}}\)

\(\Leftrightarrow A=\dfrac{2}{2\sqrt{2}}+\dfrac{2}{2\sqrt{3}}+....+\dfrac{2}{2\sqrt{100}}\)

\(\Leftrightarrow A=\dfrac{2}{\sqrt{2}+\sqrt{2}}+\dfrac{2}{\sqrt{3}+\sqrt{3}}+....+\dfrac{2}{\sqrt{99}+\sqrt{99}}+\dfrac{2}{\sqrt{100}+\sqrt{100}}\)

\(\Leftrightarrow A=2\left(\dfrac{1}{\sqrt{2}+\sqrt{2}}+\dfrac{1}{\sqrt{3}+\sqrt{3}}+...+\dfrac{1}{\sqrt{99}+\sqrt{99}}+\dfrac{1}{\sqrt{100}+\sqrt{100}}\right)\)

Ta có:

\(\dfrac{1}{\sqrt{2}+\sqrt{2}}< \dfrac{1}{1+\sqrt{2}};\dfrac{1}{\sqrt{3}+\sqrt{3}}< \dfrac{1}{\sqrt{2}+\sqrt{3}}\)

Tường tự, ta có:

\(\dfrac{A}{2}< \dfrac{1}{1+\sqrt{2}}+\dfrac{1}{\sqrt{2}+\sqrt{3}}+...+\dfrac{1}{\sqrt{99}+\sqrt{100}}\)

\(A< 2\left(\dfrac{1-\sqrt{2}}{-1}+\dfrac{\sqrt{2}-\sqrt{3}}{-1}+\dfrac{\sqrt{99}-\sqrt{100}}{-1}\right)\)

\(A< -2\left(1-\sqrt{2}+\sqrt{2}-\sqrt{3}+...-\sqrt{99}+\sqrt{99}-\sqrt{100}\right)\)

\(A< -2\left(1-\sqrt{100}\right)\)

\(A< 18\)

Vậy\(\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{3}}+\dfrac{1}{\sqrt{4}}+...+\dfrac{1}{\sqrt{100}}< 18\)

ta có :

\(\dfrac{1}{\sqrt{1}}+\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{3}}+...+\dfrac{1}{\sqrt{100}}\)

\(=\dfrac{2}{2\sqrt{1}}+\dfrac{2}{2\sqrt{2}}+\dfrac{2}{2\sqrt{3}}+...+\dfrac{2}{2\sqrt{100}}\)

\(>\dfrac{2}{\sqrt{1}+\sqrt{2}}+\dfrac{2}{\sqrt{2}+\sqrt{3}}+\dfrac{2}{\sqrt{3}+\sqrt{4}}+...+\dfrac{2}{\sqrt{100}+\sqrt{101}}\)

\(=2\left(\dfrac{1}{\sqrt{1}+\sqrt{2}}+\dfrac{1}{\sqrt{2}+\sqrt{3}}+\dfrac{1}{\sqrt{3}+\sqrt{4}}+...+\dfrac{1}{\sqrt{100}+\sqrt{101}}\right)\)

\(=2\left(\dfrac{\sqrt{1}-\sqrt{2}}{1-2}+\dfrac{\sqrt{2}-\sqrt{3}}{2-3}+\dfrac{\sqrt{3}+\sqrt{4}}{3-4}+...+\dfrac{\sqrt{100}-\sqrt{101}}{100-101}\right)\)

\(=2\left(\dfrac{\sqrt{1}-\sqrt{101}}{-1}\right)=2\left(\sqrt{101}-\sqrt{1}\right)=18,1\)

\(>18\)

\(\Rightarrow\dfrac{1}{\sqrt{1}}+\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{3}}+...+\dfrac{1}{\sqrt{100}}>18\)

10 chứ.Nếu là 10 tui làm cho nhé

\(A=\dfrac{1}{\sqrt{1}+\sqrt{2}}+\dfrac{1}{\sqrt{2}+\sqrt{3}}+...+\dfrac{1}{\sqrt{99}+\sqrt{100}}\)

\(=\dfrac{\sqrt{2}-\sqrt{1}}{\left(\sqrt{1}+\sqrt{2}\right)\left(\sqrt{2}-\sqrt{1}\right)}+\dfrac{\sqrt{3}-\sqrt{2}}{\left(\sqrt{3}-\sqrt{2}\right)\left(\sqrt{3}+\sqrt{2}\right)}+...+\dfrac{\sqrt{100}-\sqrt{99}}{\left(\sqrt{100}-\sqrt{99}\right)\left(\sqrt{100}+\sqrt{99}\right)}\)

\(=\sqrt{2}-\sqrt{1}+\sqrt{3}-\sqrt{2}+...+\sqrt{100}-\sqrt{99}=\sqrt{100}-\sqrt{1}=10-1=9\)

cả 2 ý bạn trục căn thức ở mấu là xong nhé:

vd: \(\dfrac{1}{\sqrt{1}+\sqrt{2}}=\dfrac{\sqrt{1}-\sqrt{2}}{-1}\). Rồi tương tự như vậy

\(\dfrac{1}{\sqrt{1}}+\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{3}}+...+\dfrac{1}{\sqrt{100}}>\dfrac{1}{\sqrt{100}}+\dfrac{1}{\sqrt{100}}+\dfrac{1}{\sqrt{100}}+...+\dfrac{1}{\sqrt{100}}\)

(100 số số hạng)

\(\Rightarrow\dfrac{1}{\sqrt{1}}+\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{3}}+...+\dfrac{1}{\sqrt{100}}>\dfrac{100}{\sqrt{100}}=\dfrac{100}{10}=10\)

Ta có:

\(\dfrac{1}{\sqrt{1}}>\dfrac{1}{10}\)

\(\dfrac{1}{\sqrt{2}}>\dfrac{1}{10}\)

\(\dfrac{1}{\sqrt{3}}>\dfrac{1}{10}\)

...

\(\dfrac{1}{\sqrt{100}}=\dfrac{1}{10}\)

\(\Rightarrow\dfrac{1}{\sqrt{1}}+\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{3}}+...+\dfrac{1}{\sqrt{100}}>100.\dfrac{1}{10}=10\).

a) Ta có: \(\left(\dfrac{2}{\sqrt{3}-1}+\dfrac{3}{\sqrt{3}-2}+\dfrac{15}{3-\sqrt{3}}\right)\cdot\dfrac{1}{5+\sqrt{3}}\)

\(=\left(\dfrac{2\left(\sqrt{3}+1\right)}{2}-\dfrac{3\left(2+\sqrt{3}\right)}{\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)}+\dfrac{15\left(3+\sqrt{3}\right)}{6}\right)\cdot\dfrac{1}{5+\sqrt{3}}\)

\(=\left(\sqrt{3}+1-6-3\sqrt{3}+\dfrac{15}{2}+\dfrac{5}{2}\sqrt{3}\right)\cdot\dfrac{1}{5+\sqrt{3}}\)

\(=\left(\dfrac{5}{2}+\dfrac{\sqrt{3}}{2}\right)\cdot\dfrac{1}{5+\sqrt{3}}\)

\(=\dfrac{1}{2}\)

b) Ta có: \(\dfrac{1}{\sqrt{2}+1}+\dfrac{1}{\sqrt{3}+\sqrt{2}}+...+\dfrac{1}{\sqrt{100}+\sqrt{99}}\)

\(=-1+\sqrt{2}-\sqrt{2}+\sqrt{3}-...-\sqrt{99}+\sqrt{100}\)

=-1+10=9

Với n\(\in N\)^* có: \(\dfrac{1}{\left(n+1\right)\sqrt{n}+n\sqrt{n+1}}=\dfrac{1}{\sqrt{n\left(n+1\right)}\left(\sqrt{n+1}+\sqrt{n}\right)}\)\(=\dfrac{\sqrt{n+1}-\sqrt{n}}{\sqrt{n\left(n+1\right)}\left(n+1-n\right)}=\dfrac{\sqrt{n+1}-\sqrt{n}}{\sqrt{n\left(n+1\right)}}\)\(=\dfrac{1}{\sqrt{n}}-\dfrac{1}{\sqrt{n+1}}\)

\(\Rightarrow\)\(\dfrac{1}{\left(n+1\right)\sqrt{n}+n\sqrt{n+1}}=\dfrac{1}{\sqrt{n}}-\dfrac{1}{\sqrt{n+1}}\) (*)

a) Áp dụng (*) vào T

\(\Rightarrow T=1-\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{2}}-\dfrac{1}{\sqrt{3}}+...+\dfrac{1}{\sqrt{99}}-\dfrac{1}{\sqrt{100}}\)\(=1-\dfrac{1}{10}=\dfrac{9}{10}\)

b) Có \(VT=1-\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{2}}-\dfrac{1}{\sqrt{3}}+...+\dfrac{1}{\sqrt{n}}-\dfrac{1}{\sqrt{n+1}}\)\(=1-\dfrac{1}{\sqrt{n+1}}=\dfrac{4}{5}\)

\(\Leftrightarrow\sqrt{n+1}=5\Leftrightarrow n=24\) (tm)

Vậy n=24.