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\(A=\frac{2}{\sqrt{3+\frac{2}{\sqrt{3+\frac{2}{\sqrt{3+\frac{2}{\sqrt{3+\frac{2}{\sqrt{3+1}}}}}}}}}}=\frac{2}{\sqrt{3+\frac{2}{\sqrt{3+\frac{2}{\sqrt{3+\frac{2}{\sqrt{3+1}}}}}}}}\)
\(=\frac{2}{\sqrt{3+\frac{2}{\sqrt{3+\frac{2}{\sqrt{3+1}}}}}}=\frac{2}{\sqrt{3+\frac{2}{\sqrt{3+1}}}}=\frac{2}{\sqrt{3+1}}=1\)
Ta có :
\(A=\frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{4}}+...+\frac{1}{\sqrt{n-1}+\sqrt{n}}\)
Ta có:
\(\frac{1}{\sqrt{x}+\sqrt{x-1}}=\frac{\sqrt{x}-\sqrt{x-1}}{\left(\sqrt{x}+\sqrt{x-1}\right)\left(\sqrt{x}-\sqrt{x-1}\right)}=\sqrt{x}-\sqrt{x-1}\)
Do đó:
\(A=\frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{4}}+...+\frac{1}{\sqrt{n-1}+\sqrt{n}}\)
\(\Leftrightarrow A=\sqrt{1}-\sqrt{2}+\sqrt{2}-\sqrt{3}+\sqrt{3}-\sqrt{4}+...+\sqrt{n-1}+\sqrt{n}\)
\(\Leftrightarrow A=\sqrt{n}-1\left(dpcm\right)\)
Ta có: \(\frac{1}{n\sqrt{n+1}+\left(n+1\right)\sqrt{n}}=\frac{1}{\sqrt{n}.\sqrt{n+1}\left(\sqrt{n}+\sqrt{n+1}\right)}\)
\(=\frac{\sqrt{n+1}-\sqrt{n}}{\sqrt{n}.\sqrt{n+1}\left(\sqrt{n}+\sqrt{n+1}\right)\left(\sqrt{n+1}-\sqrt{n}\right)}=\frac{\sqrt{n+1}-\sqrt{n}}{\sqrt{n}.\sqrt{n+1}}\)
\(=\frac{\sqrt{n+1}}{\sqrt{n}.\sqrt{n+1}}-\frac{\sqrt{n}}{\sqrt{n}.\sqrt{n+1}}=\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\)
Thay n = 1, 2, 3, ..., 2011 vào C ta có:
\(C=1-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{2011}}-\frac{1}{\sqrt{2012}}=1-\frac{1}{\sqrt{2012}}\)
Vậy \(C=1-\frac{1}{\sqrt{2012}}.\)
\(\frac{8+2\sqrt{2}}{3-\sqrt{2}}-\frac{2+3\sqrt{2}}{\sqrt{2}}+\frac{\sqrt{2}}{1-\sqrt{2}}\)
\(=\frac{\left(2+\sqrt{2}\right)\left(2^2-2\sqrt{2}+\sqrt{2}^2\right)}{3-\sqrt{2}}-\frac{\sqrt{2}\left(3+\sqrt{2}\right)}{\sqrt{2}}+\frac{\sqrt{2}}{1-\sqrt{2}}\)
\(=\frac{2\sqrt{2}\left(\sqrt{2}+1\right)\left(3-\sqrt{2}\right)}{3-\sqrt{2}}-3-\sqrt{2}\frac{\sqrt{2}\left(\sqrt{2}+1\right)}{\left(\sqrt{2}-1\right)\left(\sqrt{2}+1\right)}\)
\(=4+2\sqrt{2}-3-\sqrt{2}-2-\sqrt{2}=-1\)