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.

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

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

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

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

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

\(\Leftrightarrow x=-9\)

rút gọn biểu thức:

ta có dạng tổng quát : \($\frac{1} { $\sqrt{a}\)

VT tương đương với \(\dfrac{1}{\sqrt{1}+\sqrt{2}}+\dfrac{1}{\sqrt{2}+\sqrt{3}}+...+\dfrac{1}{\sqrt{99}+\sqrt{100}}\)

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

\(=\sqrt{100}-\sqrt{99}+\sqrt{99}-....-\sqrt{3}+\sqrt{3}-\sqrt{2}+\sqrt{2}-\sqrt{1}\) (kiểu do mẫu số nó có kết quả âm nên đảo lại phép)

\(=10-1=9=VP\)

\(\forall n\in N\) ta luôn có \(\dfrac{1}{\sqrt{n}+\sqrt{n+1}}=\sqrt{n+1}-\sqrt{n}\) (*)

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

\(\Leftrightarrow\left(n+1\right)-n=1\) (luôn đúng)

Vậy (*) được chứng minh.

Áp dụng với \(n=1;2;3;...;99\) ta có

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

\(=\sqrt{2}-1+\sqrt{3}-\sqrt{2}+\sqrt{4}-\sqrt{3}+...+\sqrt{100}-\sqrt{99}\)

\(=\sqrt{100}-1=10-1=9\)

Vậy S là 1 số nguyên.

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

Ta có:

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

\(\dfrac{1}{\sqrt{2}}>\dfrac{1}{\sqrt{100}}=\dfrac{1}{10}\)

\(...............\)

\(\dfrac{1}{\sqrt{98}}>\dfrac{1}{\sqrt{100}}=\dfrac{1}{10}\)

\(\dfrac{1}{\sqrt{99}}>\dfrac{1}{\sqrt{100}}=\dfrac{1}{10}\)

Cộng theo vế ta có:

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

Lại có \(\dfrac{1}{\sqrt{100}}=\dfrac{1}{10}\) suy ra:

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

Ta có:

1/√1>1/√100=1/10

1/√2>1/√100=1/10

........

1/√100=1/√100=1/10

Nên:

1/√1+1/√2+...+1/√100>1/10+1/10+...+1/10(100 phân số 1/10)

=1/√1+1/√2+..+1/√100>100/10

1/√1+1/√2+..+1/√100>10(đpcm)

Ta có:

Cộng theo vế ta có:

Lại có suy ra: