Dạng tổng quát: Với n là các số lẻ lớn hơn hoặc bằng 3 thì \(\frac{1}{n\sqrt{n-2}+\left(n-2\right)\sqrt{n}}=\frac{1}{\sqrt{n\left(n-2\right)}\left(\sqrt{n}+\sqrt{n-2}\right)}=\frac{1}{\sqrt{n\left(n-2\right)}.\frac{2}{\sqrt{n}-\sqrt{n-2}}}=\frac{\sqrt{n}-\sqrt{n-2}}{2\sqrt{n\left(n-2\right)}}=\frac{1}{2}\left(\frac{1}{\sqrt{n-2}}-\frac{1}{\sqrt{n}}\right)\)Áp dụng, ta được: \(C=\frac{1}{3\sqrt{1}+1\sqrt{3}}+\frac{1}{5\sqrt{3}+3\sqrt{5}}+...+\frac{1}{121\sqrt{119}+119\sqrt{121}}=\frac{1}{2}\left(\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{3}}+\frac{1}{\sqrt{3}}-\frac{1}{\sqrt{5}}+...+\frac{1}{\sqrt{119}}-\frac{1}{\sqrt{121}}\right)=\frac{1}{2}\left(1-\frac{1}{11}\right)=\frac{5}{11}\)Vậy C = ⁵/₁₁

Xét :\(\frac{1}{\left(a+2\right)\sqrt{a}+a\sqrt{a+2}}=\frac{1}{\sqrt{a}.\sqrt{a+2}\left(\sqrt{a+2}+\sqrt{a}\right)}=\frac{\sqrt{a+2}-\sqrt{a}}{2\sqrt{a}.\sqrt{a+2}}=\frac{1}{2\sqrt{a}}-\frac{1}{2\sqrt{a+2}}\)

Xét: 

\(C=\frac{1}{3\sqrt{1}+1\sqrt{3}}+\frac{1}{5\sqrt{3}+3\sqrt{5}}+...+\frac{1}{121\sqrt{119}+119\sqrt{121}}\)

\(=\frac{1}{2}-\frac{1}{2\sqrt{3}}+\frac{1}{2\sqrt{3}}-\frac{1}{2\sqrt{5}}+\frac{1}{2\sqrt{5}}-\frac{1}{2\sqrt{7}}+...+\frac{1}{2\sqrt{119}}-\frac{1}{2\sqrt{121}}\)

\(=\frac{1}{2}-\frac{1}{2\sqrt{121}}=\frac{1}{2}-\frac{1}{2.11}=\frac{5}{11}\)

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

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

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

Từ đó ta thay vào:

\(C=\frac{1}{2}\cdot\left(1-\frac{1}{\sqrt{3}}+\frac{1}{\sqrt{3}}-\frac{1}{\sqrt{5}}+...+\frac{1}{\sqrt{199}}-\frac{1}{\sqrt{121}}\right)\)

\(C=\frac{1}{2}\cdot\left(1-\frac{1}{11}\right)\)

\(C=\frac{1}{2}\cdot\frac{10}{11}=\frac{5}{11}\)

Vậy C = 5/11

Dạng tổng quát ta càn chứng minh \(\sqrt{\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{\left(a+b\right)^2}}=\frac{1}{a}+\frac{1}{b}-\frac{1}{a+b}\)

Ta có \(\sqrt{\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{\left(a+b\right)^2}}\)

\(=\sqrt{\frac{a^4+2a^3b+a^2b^2+2ab^3+b^4}{a^2b^2\left(a+b\right)^2}}\)

\(=\sqrt{\left(\frac{a^2+ab+b^2}{ab\left(a+b\right)}\right)^2}\)

\(=\frac{a^2+ab+b^2}{ab\left(a+b\right)}=\frac{1}{b}+\frac{b}{a\left(a+b\right)}=\frac{1}{b}+\frac{1}{a}-\frac{1}{a+b}\left(đpcm\right)\)

Áp dụng dạng trên ta được 

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

\(D=100-\frac{1}{100}=\frac{9999}{100}\)

Xét biểu thức \(A=\sqrt{1+\frac{1}{a^2}+\frac{1}{\left(a+1\right)^2}}\)với a > 0

\(A^2=1+\frac{1}{a^2}+\frac{1}{\left(a+1\right)^2}=\frac{a^2\left(a+1\right)^2+\left(a+1\right)^2+a^2}{a^2\left(a+1\right)^2}=\frac{a^2\left(a^2+2a+1+1\right)+\left(a+1\right)^2}{a^2\left(a+1\right)^2}=\frac{a^4+2a^2\left(a+1\right)+\left(a+1\right)^2}{a^2\left(a+1\right)^2}=\frac{\left(a^2+a+1\right)^2}{a^2\left(a+1\right)^2}=\left[\frac{a^2+a+1}{a\left(a+1\right)}\right]^2\)Do a > 0 nên A > 0 và \(A=\frac{a^2+a+1}{a\left(a+1\right)}=1+\frac{1}{a\left(a+1\right)}=1+\frac{1}{a}-\frac{1}{a+1}\)

Do đó \(D=\left(1+\frac{1}{1}-\frac{1}{2}\right)+\left(1+\frac{1}{2}-\frac{1}{3}\right)+\left(1+\frac{1}{3}-\frac{1}{4}\right)+...+\left(1+\frac{1}{99}-\frac{1}{100}\right)=99+\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\right)=100-\frac{1}{100}=99,99\)

a) Trục căn thức ở mỗi số hạng của biểu thức A,ta có:

 \(A=\frac{1}{\sqrt{1}-\sqrt{2}}-\frac{1}{\sqrt{2}-\sqrt{3}}+\frac{1}{\sqrt{3}-\sqrt{4}}-...+\frac{1}{\sqrt{2007}-\sqrt{2008}}\)=\(\frac{\sqrt{2}+\sqrt{1}}{1-2}-\frac{\sqrt{3}+\sqrt{2}}{2-3}+\frac{\sqrt{3}+\sqrt{4}}{3-4}-...+\frac{\sqrt{2007}+\sqrt{2008}}{2007-2008}\)

= \(-\left(\sqrt{1}+\sqrt{2}\right)+\left(\sqrt{2}+\sqrt{3}\right)-\left(\sqrt{3}+\sqrt{4}\right)+...-\left(\sqrt{2007}+\sqrt{2008}\right)\)

=\(-1-\sqrt{2008}\)

b)Ta xét số hạng tổng quát: \(\frac{1}{\left(n+1\right)\sqrt{n}+n\sqrt{n+1}}\)=\(\frac{\left(n+1\right)\sqrt{n}-n\sqrt{n+1}}{\left(n+1\right)^2n-n^2\left(n+1\right)}\)=\(\frac{\left(n+1\right)\sqrt{n}-n\sqrt{n+1}}{n\left(n+1\right)}\)=\(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\)

Áp dụng vào biểu thức B ta được: 

B= \(\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}-...+\frac{1}{\sqrt{120}}-\frac{1}{\sqrt{121}}=1-\frac{1}{11}\)= \(\frac{10}{11}\)

\(A=\frac{1}{\sqrt{1}-\sqrt{2}}-\frac{1}{\sqrt{2}-\sqrt{3}}+\frac{1}{\sqrt{3}-\sqrt{4}}-\frac{1}{\sqrt{4}-\sqrt{5}}+...+\frac{1}{\sqrt{2007}-\sqrt{2008}}\)

\(=\frac{-1}{\sqrt{2}-\sqrt{1}}+\frac{1}{\sqrt{3}-\sqrt{2}}-\frac{1}{\sqrt{4}-\sqrt{3}}+\frac{1}{\sqrt{5}-\sqrt{4}}-....+\frac{1}{\sqrt{2007}-\sqrt{2006}}-\frac{1}{\sqrt{2008}-\sqrt{2007}}\)

\(=\frac{-1\cdot\left(\sqrt{2}+\sqrt{1}\right)}{2-1}+\frac{1\cdot\left(\sqrt{3}+\sqrt{2}\right)}{3-2}-\frac{1\cdot\left(\sqrt{4}+\sqrt{3}\right)}{4-3}+\frac{1\cdot\left(\sqrt{5}+\sqrt{4}\right)}{5-4}-...+\frac{1\cdot\left(\sqrt{2007}+\sqrt{2006}\right)}{2007-2006}-\frac{1 \left(\sqrt{2008}+\sqrt{2007}\right)}{2008-2007}\)

\(=-1-\sqrt{2}+\sqrt{2}+\sqrt{3}-\sqrt{3}-\sqrt{4}+\sqrt{4}+\sqrt{5}-...+\sqrt{2006}+\sqrt{2007}-\sqrt{2007}-\sqrt{2008}\) 

\(=-1-\sqrt{2008}\)

Với \(k\in N;k\ne0\) ta có :

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

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

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

Áp dụng ta có :

\(M=\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+....+\frac{1}{\sqrt{120}}-\frac{1}{\sqrt{121}}=1-\frac{1}{11}=\frac{10}{11}\)

Xét \(A=\sqrt{1+\frac{1}{a^2}+\frac{1}{\left(a+1\right)^2}}a>0\)

Ta có: \(A^2=1+\frac{1}{a^2}+\frac{1}{\left(a+1\right)^2}=\frac{a^2\left(a+1\right)^2+\left(a+1\right)^2+a^2}{a^2\left(a+1\right)^2}\)

\(\frac{a^4+2a^2\left(a+1\right)+\left(a+1\right)^2}{a^2\left(a+1\right)^2}=\frac{\left(a^2+a+1\right)^2}{a^2\left(a+1\right)^2}\)

Vì a>0, D>0  nên \(A=\frac{a^2+a+1}{a\left(a+1\right)}=1+\frac{1}{a}-\frac{1}{a+1}\)

Áp dụng ta có: \(D=\sqrt{1+\frac{1}{1^2}+\frac{1}{2^2}}+\sqrt{1+\frac{1}{2^2}+\frac{1}{3^2}}+...+\sqrt{1+\frac{1}{99^2}+\frac{1}{100^2}}\)

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