\(A=\frac{\sqrt{2}-1}{\left(\sqrt{2}-1\right)\left(\sqrt{2}+1\right)}+\frac{\sqrt{3}-\sqrt{2}}{\left(\sqrt{3}-\sqrt{2}\right)\left(\sqrt{3}+\sqrt{2}\right)}+...+\frac{\sqrt{121}-\sqrt{120}}{\left(\sqrt{121}-\sqrt{120}\right)\left(\sqrt{121}+\sqrt{120}\right)}\)

\(=\sqrt{2}-1+\sqrt{3}-\sqrt{2}+...+\sqrt{121}-\sqrt{120}\)

\(=\sqrt{121}-1=10\)

\(B=\frac{2}{2.\sqrt{1}}+\frac{2}{2\sqrt{2}}+\frac{2}{2\sqrt{3}}+...+\frac{2}{2\sqrt{35}}\)

\(B>\frac{1}{1+\sqrt{2}}+\frac{2}{\sqrt{2}+\sqrt{3}}+\frac{2}{\sqrt{3}+\sqrt{4}}+...+\frac{2}{\sqrt{35}+\sqrt{36}}\)

\(B>2\left(\frac{\sqrt{2}-\sqrt{1}}{\left(\sqrt{2}-\sqrt{1}\right)\left(\sqrt{2}+\sqrt{1}\right)}+...+\frac{\sqrt{36}-\sqrt{35}}{\left(\sqrt{36}-\sqrt{35}\right)\left(\sqrt{36}+\sqrt{35}\right)}\right)\)

\(B>2\left(\sqrt{36}-\sqrt{1}\right)=10\Rightarrow B>A\)

Bài 2:

\(D=\frac{1}{2\sqrt{1}+1\sqrt{2}}+\frac{1}{3\sqrt{2}+2\sqrt{3}}+\frac{1}{4\sqrt{3}+3\sqrt{4}}+...+\frac{1}{120\sqrt{121}+121\sqrt{120}}\)

Với mọi \(n\inℕ^∗\)ta có:

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

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

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

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

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

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

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

\(\Rightarrow D=\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+\frac{1}{\sqrt{3}}-\frac{1}{\sqrt{4}}+\frac{1}{\sqrt{4}}+....+\frac{1}{\sqrt{120}}-\frac{1}{\sqrt{121}}\)

\(=1-\frac{1}{\sqrt{121}}=\frac{10}{11}\)

Bài 1: chắc lại phải "liên hợp" gì đó rồi:V

\(\sqrt{2009}-\sqrt{2008}=\frac{1}{\sqrt{2009}+\sqrt{2008}}\)

\(\sqrt{2007}-\sqrt{2006}=\frac{1}{\sqrt{2007}+\sqrt{2006}}\)

Đó \(\sqrt{2009}+\sqrt{2008}>\sqrt{2007}+\sqrt{2006}\)

Nên \(\sqrt{2009}-\sqrt{2008}< \sqrt{2007}-\sqrt{2006}\)

Tổng quát ta có bài toán sau, với So sánh \(\sqrt{n}-\sqrt{n-1}\text{ và }\sqrt{n-2}-\sqrt{n-3}\)

Với \(n\ge3\). Lời giải xin mời các bạn:)

Câu a) 

Có: \(A=\sqrt{2009}-\sqrt{2008}\Leftrightarrow A^2=1-2\sqrt{2009\cdot2008}\)

\(B=\sqrt{2007}-\sqrt{2006}\Rightarrow B^2=1-2\sqrt{2007\cdot2006}\)

Đương nhiên: \(2\sqrt{2009\cdot2008}>2\sqrt{2006\cdot2007}\)

Suy ra: \(A< B\)

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

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

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

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

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

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

Ta trục căn thức ở mỗi số hạng của A sau đó khử liên tiếp đc : A = 11 - 1 = 10

Ta có : \(B=\frac{2}{2\sqrt{1}}+\frac{2}{2\sqrt{2}}+...+\frac{2}{2\sqrt{35}}\)

\(B=\frac{2}{\sqrt{1}+\sqrt{1}}+\frac{2}{\sqrt{2}+\sqrt{2}}+...+\frac{2}{\sqrt{35}+\sqrt{35}}\)

\(B>2\left(\frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+...+\frac{1}{\sqrt{35}+\sqrt{36}}\right)\)

\(B>2\left(\sqrt{2}-\sqrt{1}+\sqrt{3}-\sqrt{2}+...+\sqrt{36}-\sqrt{35}\right)\)

\(B>2\left(6-1\right)=10\)

Vậy A < B