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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:
\(A=\frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{4}}+...+\frac{1}{\sqrt{n-1}+\sqrt{n}}\)
\(\Leftrightarrow A=\sqrt{1}-\sqrt{2}+\sqrt{2}-\sqrt{3}+\sqrt{3}-\sqrt{4}+...+\sqrt{n-1}+\sqrt{n}\)
\(\Leftrightarrow A=\sqrt{n}-1\left(dpcm\right)\)
\(\frac{2+\sqrt{3}}{\sqrt{2}+\sqrt{2}+\sqrt{3}}+\frac{2-\sqrt{3}}{\sqrt{2}+\sqrt{2}-\sqrt{3}}\)
= \(\frac{2+\sqrt{3}}{2\sqrt{2}+\sqrt{3}}+\frac{2-\sqrt{3}}{2\sqrt{2}-\sqrt{3}}\)
= \(\frac{\left(2+\sqrt{3}\right)\left(2\sqrt{2}-\sqrt{3}\right)+\left(2-\sqrt{3}\right)\left(2\sqrt{2}+\sqrt{3}\right)}{\left(2\sqrt{2}+\sqrt{3}\right)\left(2\sqrt{2}-\sqrt{3}\right)}\)
= \(\frac{4\sqrt{2}-2\sqrt{3}+2\sqrt{6}-3+4\sqrt{2}+2\sqrt{3}-2\sqrt{6}-3}{\left(2\sqrt{2}\right)^2-\left(\sqrt{3}\right)^2}\)
= \(\frac{8\sqrt{2}-6}{-5}\)
Chúc bạn học tốt !!!
\(A=\frac{2}{\sqrt{3+\frac{2}{\sqrt{3+\frac{2}{\sqrt{3+\frac{2}{\sqrt{3+\frac{2}{\sqrt{3+1}}}}}}}}}}=\frac{2}{\sqrt{3+\frac{2}{\sqrt{3+\frac{2}{\sqrt{3+\frac{2}{\sqrt{3+1}}}}}}}}\)
\(=\frac{2}{\sqrt{3+\frac{2}{\sqrt{3+\frac{2}{\sqrt{3+1}}}}}}=\frac{2}{\sqrt{3+\frac{2}{\sqrt{3+1}}}}=\frac{2}{\sqrt{3+1}}=1\)
\(\frac{8+2\sqrt{2}}{3-\sqrt{2}}-\frac{2+3\sqrt{2}}{\sqrt{2}}+\frac{\sqrt{2}}{1-\sqrt{2}}\)
\(=\frac{\left(2+\sqrt{2}\right)\left(2^2-2\sqrt{2}+\sqrt{2}^2\right)}{3-\sqrt{2}}-\frac{\sqrt{2}\left(3+\sqrt{2}\right)}{\sqrt{2}}+\frac{\sqrt{2}}{1-\sqrt{2}}\)
\(=\frac{2\sqrt{2}\left(\sqrt{2}+1\right)\left(3-\sqrt{2}\right)}{3-\sqrt{2}}-3-\sqrt{2}\frac{\sqrt{2}\left(\sqrt{2}+1\right)}{\left(\sqrt{2}-1\right)\left(\sqrt{2}+1\right)}\)
\(=4+2\sqrt{2}-3-\sqrt{2}-2-\sqrt{2}=-1\)
\(\frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}}+\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}\)
\(=\frac{\left(\sqrt{3}-\sqrt{2}\right)^2+\left(\sqrt{3}+\sqrt{2}\right)^2}{\left(\sqrt{3}+\sqrt{2}\right)\left(\sqrt{3}-\sqrt{2}\right)}\)
\(=\frac{3-2\sqrt{3}.\sqrt{2}+3+2\sqrt{3}.\sqrt{2}+2}{3-2}\)
\(=3+2+3+2=10\)