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Ta có
\(\frac{1}{\left(n+1\right)\sqrt{n}+n\sqrt{n+1}}=\frac{\left(n+1\right)\sqrt{n}-n\sqrt{n+1}}{n^2+n}\)
\(=\frac{\sqrt{n+1}-\sqrt{n}}{\sqrt{n+1}\sqrt{n}}=\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\)
Từ đó ta có
\(A=\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{2}}+...+\frac{1}{\sqrt{2004}}-\frac{1}{\sqrt{2005}}\)
\(=1-\frac{1}{\sqrt{2005}}=\frac{\sqrt{2005}-1}{\sqrt{2005}}\)
1/ \(\sqrt{\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}}=\left|\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right|\)
\(\Leftrightarrow\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}=\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+2\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\right)\)
\(\Leftrightarrow\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}=0\)
\(\Leftrightarrow\frac{a+b+c}{abc}=0\)(đúng)
Vậy ta có ĐPCM
2/ \(A=\frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+...+\frac{1}{\sqrt{2005}+\sqrt{2006}}\)
\(=\sqrt{2}-\sqrt{1}+\sqrt{3}-\sqrt{2}+...+\sqrt{2006}-\sqrt{2005}\)
\(=\sqrt{2006}-1\)
B=\(\frac{6-6\sqrt{3}}{1-\sqrt{3}}+\frac{3\sqrt{3}+3}{\sqrt{3}+1}=\frac{6\left(1-\sqrt{3}\right)}{1-\sqrt{3}}+\frac{3\left(\sqrt{3}+1\right)}{\sqrt{3}+1}=6+3=9\)
C=\(\frac{3+\sqrt{3}}{\sqrt{3}}+\frac{\sqrt{6}-\sqrt{3}}{1-\sqrt{2}}=\frac{3\left(1+\sqrt{3}\right)}{\sqrt{3}}+\frac{\sqrt{3}\left(\sqrt{2}-1\right)}{1-\sqrt{2}}=\sqrt{3}+1-\sqrt{3}=1\)
D=\(\frac{\sqrt{10}-\sqrt{2}}{\sqrt{5}-1}+\frac{2-\sqrt{2}}{\sqrt{2}-1}=\frac{\sqrt{2}\left(\sqrt{5}-1\right)}{\sqrt{5}-1}+\frac{\sqrt{2}\left(\sqrt{2}-1\right)}{\sqrt{2}-1}=\sqrt{2}+\sqrt{2}=2\sqrt{2}\)
E=\(\frac{\sqrt{15}-\sqrt{12}}{\sqrt{5}-2}+\frac{1}{2-\sqrt{3}}=\frac{\sqrt{3}\left(\sqrt{5}-2\right)}{\sqrt{5}-2}+\frac{1}{2-\sqrt{3}}=\sqrt{3}+\frac{1}{2-\sqrt{3}}=\frac{2\sqrt{3}-1}{2-\sqrt{3}}\)
1. Trục căn thức ở mẫu:
\(A=\frac{1}{1+\sqrt{5}}+\frac{1}{\sqrt{5}+\sqrt{9}}+\frac{1}{\sqrt{9}+\sqrt{13}}+....+\frac{1}{\sqrt{2001}+\sqrt{2005}}+\frac{1}{\sqrt{2005}+\sqrt{2009}}\)
=\(\frac{\sqrt{5}-1}{4}+\frac{\sqrt{9}-\sqrt{5}}{4}+\frac{\sqrt{13}-\sqrt{9}}{4}+....+\frac{\sqrt{2005}-\sqrt{2001}}{4}+\frac{\sqrt{2009}-\sqrt{2005}}{4}\)
\(=\frac{\sqrt{2009}-1}{4}\)
2/ \(x=\sqrt[3]{3+2\sqrt{2}}+\sqrt[3]{3-2\sqrt{2}}\)
=> \(x^3=\left(\sqrt[3]{3+2\sqrt{2}}+\sqrt[3]{3-2\sqrt{2}}\right)^3\)
\(=3+2\sqrt{2}+3-2\sqrt{2}+3\left(\sqrt[3]{3+2\sqrt{2}}+\sqrt[3]{3-2\sqrt{2}}\right).\sqrt[3]{3+2\sqrt{2}}.\sqrt[3]{3-2\sqrt{2}}\)
\(=6+3x\)
=> \(x^3-3x=6\)
=> \(B=x^3-3x+2000=6+2000=2006\)
\(A=\frac{1-\sqrt{5}}{1-5}+\frac{\sqrt{5}-\sqrt{9}}{5-9}+\frac{\sqrt{9}-\sqrt{13}}{9-13}+...+\frac{\sqrt{2001}-\sqrt{2005}}{2001-2005}\)
\(A=\frac{1-\sqrt{5}+\sqrt{5}-\sqrt{9}+\sqrt{9}-\sqrt{13}+...+\sqrt{2001}-\sqrt{2005}}{-4}\)
\(A=\frac{1-\sqrt{2005}}{-4}=\frac{\sqrt{2005}-1}{4}\)
Ap dung cong thuc \(\sqrt{1+\frac{1}{a^2}+\frac{1}{\left(a+1\right)^2}}=1+\frac{1}{a}-\frac{1}{a+1}\)
ta co \(E=1+\frac{1}{2}-\frac{1}{3}+1+\frac{1}{3}-\frac{1}{4}+...+1+\frac{1}{2005}-\frac{1}{2006}=2004+\frac{1}{2}-\frac{1}{2006}\)
Ta có:
\(E=\sqrt{\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{\left(-3\right)^2}}+\sqrt{\frac{1}{1^2}+\frac{1}{3^2}+\frac{1}{\left(-4\right)^2}}+...+\sqrt{\frac{1}{1^2}+\frac{1}{2005^2}+\frac{1}{\left(-2006\right)^2}}\)
DO: \(1+2+\left(-3\right)=0;1+3+\left(-4\right)=0;...;1+2005+\left(-2006\right)=0\)
=> TA ĐƯỢC: \(E=\sqrt{\left(\frac{1}{1}+\frac{1}{2}+\frac{1}{-3}\right)^2}+\sqrt{\left(\frac{1}{1}+\frac{1}{3}+\frac{1}{-4}\right)^2}+...+\sqrt{\left(\frac{1}{1}+\frac{1}{2005}+\frac{1}{-2006}\right)^2}\)
=> \(E=\frac{1}{1}+\frac{1}{2}-\frac{1}{3}+\frac{1}{1}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{1}+\frac{1}{2005}-\frac{1}{2006}\)
=> \(E=\left(\frac{1}{1}+\frac{1}{1}+...+\frac{1}{1}\right)+\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2005}-\frac{1}{2006}\right)\)
DO TRONG E CÓ TẤT CẢ 2004 CĂN THỨC
=> \(E=2004+\frac{1}{2}-\frac{1}{2006}=2004+\frac{501}{1003}=\frac{2010513}{1003}\)