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Bài 2:
a) \(\frac{1}{\sqrt{1}+\sqrt{2}}=\frac{2-1}{\sqrt{1}+\sqrt{2}}=\frac{\left(\sqrt{2}-\sqrt{1}\right)\left(\sqrt{2}+\sqrt{1}\right)}{\sqrt{1}+\sqrt{2}}=\sqrt{2}-\sqrt{1}\)
Tương tự ta có: \(\frac{1}{\sqrt{2}+\sqrt{3}}=\sqrt{3}-\sqrt{2}\);
\(\frac{1}{\sqrt{3}+\sqrt{4}}=\sqrt{4}-\sqrt{3}\); ............. ; \(\frac{1}{\sqrt{2024}+\sqrt{2025}}=\sqrt{2025}-\sqrt{2024}\)
\(\Rightarrow A=\sqrt{2}-\sqrt{1}+\sqrt{3}-\sqrt{2}+\sqrt{4}-\sqrt{3}+......+\sqrt{2025}-\sqrt{2024}\)
\(=\sqrt{2025}-\sqrt{1}=45-1=44\)
Bài 4:
\(M=\frac{\sqrt{3-2\sqrt{2}}}{\sqrt{17-12\sqrt{2}}}-\frac{\sqrt{3+2\sqrt{2}}}{\sqrt{17+12\sqrt{2}}}\)
\(=\frac{\sqrt{2-2\sqrt{2}+1}}{\sqrt{9-2.3.2\sqrt{2}+8}}-\frac{\sqrt{2+2\sqrt{2}+1}}{\sqrt{9+2.3.2\sqrt{2}+8}}\)
\(=\frac{\sqrt{\left(\sqrt{2}-1\right)^2}}{\sqrt{\left(3-\sqrt{8}\right)^2}}-\frac{\sqrt{\left(\sqrt{2}+1\right)^2}}{\sqrt{\left(3+\sqrt{8}\right)^2}}\)
\(=\frac{\left|\sqrt{2}-1\right|}{\left|3-\sqrt{8}\right|}-\frac{\left|\sqrt{2}+1\right|}{\left|3+\sqrt{8}\right|}=\frac{\sqrt{2}-1}{3-\sqrt{8}}-\frac{\sqrt{2}+1}{3+\sqrt{8}}\)
\(=\frac{\left(\sqrt{2}-1\right)\left(3+\sqrt{8}\right)}{\left(3-\sqrt{8}\right)\left(3+\sqrt{8}\right)}-\frac{\left(\sqrt{2}+1\right)\left(3-\sqrt{8}\right)}{\left(3+\sqrt{8}\right)\left(3-\sqrt{8}\right)}\)
\(=\left(3\sqrt{2}+\sqrt{16}-3-\sqrt{8}\right)-\left(3\sqrt{2}-\sqrt{16}+3-\sqrt{8}\right)\)
\(=3\sqrt{2}+4-3-\sqrt{8}-3\sqrt{2}+4-3+\sqrt{8}\)
\(=8-6=2\)là số tự nhiên
\(\frac{1}{n\sqrt{n-1}}=\frac{\sqrt{n-1}}{\left(n-1\right)n}=\sqrt{n-1}.\frac{1}{\left(n-1\right)n}=\sqrt{n-1}\left(\frac{1}{n-1}-\frac{1}{n}\right)\)
\(=\sqrt{n-1}\left(\frac{1}{\sqrt{n-1}}-\frac{1}{\sqrt{n}}\right)\left(\frac{1}{\sqrt{n-1}}+\frac{1}{\sqrt{n}}\right)\)
\(=\left(\frac{1}{\sqrt{n-1}}-\frac{1}{\sqrt{n}}\right)\left(1+\frac{\sqrt{n-1}}{\sqrt{n}}\right)\)
\(< \left(\frac{1}{\sqrt{n-1}}-\frac{1}{\sqrt{n}}\right)\left(1+\frac{\sqrt{n}}{\sqrt{n}}\right)=2\left(\frac{1}{\sqrt{n-1}}-\frac{1}{\sqrt{n}}\right)\)
Áp dụng vài bài toán:
\(\frac{1}{2}+\frac{1}{3\sqrt{2}}+\frac{1}{4\sqrt{3}}+...+\frac{1}{2005\sqrt{2004}}\)
\(< 2\left(1-\frac{1}{\sqrt{2}}\right)+2\left(\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}\right)+2\left(\frac{1}{\sqrt{3}}-\frac{1}{\sqrt{4}}\right)+...+2\left(\frac{1}{\sqrt{2004}}-\frac{1}{\sqrt{2005}}\right)\)
\(=2\left(1-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+\frac{1}{\sqrt{3}}-\frac{1}{\sqrt{4}}+...+\frac{1}{\sqrt{2004}}-\frac{1}{\sqrt{2005}}\right)\)
\(=2\left(1-\frac{1}{\sqrt{2005}}\right)=2-\frac{2}{\sqrt{2005}}< 2\)
Vậy \(\frac{1}{2}+\frac{1}{3\sqrt{2}}+\frac{1}{4\sqrt{3}}+...+\frac{1}{2005\sqrt{2004}}< 2\)
1.\(x=7+4\sqrt{3}\)
\(=\left(\sqrt{3}+2\right)^2\)
Thay x=\(\left(2+\sqrt{3}\right)^2\), ta có:
\(A=\frac{3+\sqrt{3}}{4+\sqrt{3}}\)
2. \(B=\frac{\sqrt{x}\left(\sqrt{x}-2\right)-\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)-\sqrt{x}-4}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}\)
\(B=\frac{-3}{2-\sqrt{x}}\left(đpcm\right)\)
3. \(\frac{B}{A}=\frac{\frac{-3}{2-\sqrt{x}}}{\frac{\sqrt{x}+1}{\sqrt{x}+2}}=\frac{-3}{2-\sqrt{x}}.\frac{\sqrt{x}+2}{\sqrt{x}+1}\)
\(\frac{B}{A}< -1\Rightarrow\frac{3\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}< -1\)
\(\Leftrightarrow\frac{3\sqrt{x}+6+x-\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}< 0\)
\(\Leftrightarrow\frac{x-2\sqrt{x}+4}{x-\sqrt{x}-2}< 0\)
\(\Rightarrow x-\sqrt{x}-2< 0\)(Vì \(x-2\sqrt{x}+4>0\))
\(\Leftrightarrow-1< x< 2\)
\(\frac{1}{2\sqrt{1}}\)+ \(\frac{1}{3\sqrt{2}}\)+...+\(\frac{1}{2008\sqrt{2007}}\)<\(\frac{88}{45}\)
