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

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

\(=\sqrt{100}-1=9\)

\(x^3+3.9x^2+3.9^2x+9^3=0\)

\(\Leftrightarrow\left(x+9\right)^3=0\)

\(\Leftrightarrow x=-9\)

Ta có: \(S=\dfrac{1}{2+\sqrt{2}}+\dfrac{1}{3\sqrt{2}+2\sqrt{3}}+\dfrac{1}{4\sqrt{3}+3\sqrt{4}}+...+\dfrac{1}{100\sqrt{99}+99\sqrt{100}}\)

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

\(=1-\dfrac{1}{10}=\dfrac{9}{10}\)

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.

Lời giải:
Xét số hạng tổng quát: 

\(\frac{\sqrt{n+1}-\sqrt{n}}{n+(n+1)}< \frac{\sqrt{n+1}-\sqrt{n}}{2\sqrt{n(n+1)}}=\frac{1}{2}(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}})\) theo BĐT Cô-si.

Do đó:
\(x< \frac{1}{2}\left[\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+....+\frac{1}{\sqrt{99}}-\frac{1}{\sqrt{100}}\right]=\frac{1}{2}(1-\frac{1}{\sqrt{100}})< \frac{1}{2}\)

Ta có đpcm.

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

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

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

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

\(=-1+10\)

\(=9\)

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

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

=10-1

=9

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

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

\(=\left(11+2\cdot\sqrt{2}+2\cdot\sqrt{3}+...+2\cdot\sqrt{99}\right)\)