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Ta có:\(\frac{1}{A}=\frac{\sqrt{a-2003}+\sqrt{b-2003}}{\sqrt{a+b}}=\frac{\sqrt{a-2003}}{\sqrt{a+b}}+\frac{\sqrt{b-2003}}{\sqrt{a+b}}\)
Mặt khác:\(\frac{1}{a}+\frac{1}{b}=\frac{1}{2003}\Rightarrow\frac{a+b}{ab}=\frac{1}{2003}\Rightarrow2003=\)\(\frac{ab}{a+b} \left(1\right)\)
Thay (1) vào \(\frac{1}{A}\) ta được: \(\frac{1}{A}=\frac{\sqrt{a-\frac{ab}{a+b}}}{\sqrt{a+b}}+\frac{\sqrt{b-\frac{ab}{a+b}}}{\sqrt{a+b}}\)
\(\Leftrightarrow\frac{1}{A}=\sqrt{\frac{a-\frac{ab}{a+b}}{a+b}}+\sqrt{\frac{b-\frac{ab}{a+b}}{a+b}}\)
\(\Leftrightarrow\frac{1}{A}=\sqrt{\frac{\frac{a^2+ab-ab}{a+b}}{a+b}}+\sqrt{\frac{\frac{b^2+ab-ab}{a+b}}{a+b}}=\sqrt{\frac{a^2}{\left(a+b\right)^2}}+\sqrt{\frac{b^2}{\left(a+b\right)^2}}\)
\(\Leftrightarrow\frac{1}{A}=\left|\frac{a}{a+b}\right|+\left|\frac{b}{a+b}\right|=\frac{a}{a+b}+\frac{b}{a+b}\left(a>2003;b>2003\right)\)
\(\Leftrightarrow\frac{1}{A}=\frac{a+b}{a+b}=1\Leftrightarrow A=1\)
Vậy............................
Dat bieu thuc tren la A
ta co \(\frac{1}{\sqrt{n+2}+\sqrt{n}}=\frac{\sqrt{n+2}-\sqrt{n}}{2}\)
ap dung dang thuc tren ta co\(\frac{1}{\sqrt{3}+1}=\frac{\sqrt{3}-1}{2}\)
tuong tu ta co \(\frac{1}{\sqrt{5}+\sqrt{3}}=\frac{\sqrt{5}-\sqrt{3}}{2}\)
.........
\(\frac{1}{\sqrt{2017}+\sqrt{2015}}=\frac{\sqrt{2017}-\sqrt{2015}}{2}\)
ta co
\(A=\frac{1}{2}\left(\sqrt{3}-1+\sqrt{5}-\sqrt{3}+.....+\sqrt{2017}-\sqrt{2015}\right)=\frac{\sqrt{2017}-1}{2}\)
Bài 2:
\(P=\frac{1}{1+\sqrt{5}}+\frac{1}{\sqrt{5}+\sqrt{9}}+...+\frac{1}{\sqrt{2001}+\sqrt{2005}}\)
\(=\frac{1-\sqrt{5}}{\left(1+\sqrt{5}\right)\left(1-\sqrt{5}\right)}+\frac{\sqrt{5}-\sqrt{9}}{\left(\sqrt{5}+\sqrt{9}\right)\left(\sqrt{5}-\sqrt{9}\right)}+...+\frac{\sqrt{2001}-\sqrt{2005}}{\left(\sqrt{2001}+\sqrt{2005}\right)\left(\sqrt{2001}-\sqrt{2005}\right)}\)
\(=\frac{1-\sqrt{5}}{1-5}+\frac{\sqrt{5}-\sqrt{9}}{5-9}+...+\frac{\sqrt{2001}-\sqrt{2005}}{2001-2005}\)
\(=\frac{1-\sqrt{5}}{-4}+\frac{\sqrt{5}-\sqrt{9}}{-4}+..+\frac{\sqrt{2001}-\sqrt{2005}}{-4}\)
\(=\frac{1-\sqrt{5}+\sqrt{5}-\sqrt{9}+...+\sqrt{2001}-\sqrt{2005}}{-4}\)
\(=\frac{1-\sqrt{2005}}{-4}\)
\(=\frac{\sqrt{2005}-1}{4}\)
\(Q=\frac{1-\sqrt{5}}{1-5}+\frac{\sqrt{5}-\sqrt{9}}{5-9}+\frac{\sqrt{9}-\sqrt{13}}{9-13}+...+\frac{\sqrt{2001}-\sqrt{2005}}{2001-2005}\)
=> \(Q=\frac{1-\sqrt{5}}{-4}+\frac{\sqrt{5}-\sqrt{9}}{-4}+\frac{\sqrt{9}-\sqrt{13}}{-4}+...+\frac{\sqrt{2001}-\sqrt{2005}}{-4}\)
=> \(Q=-\frac{1}{4}.\left(1-\sqrt{5}+\sqrt{5}-\sqrt{9}+\sqrt{9}-\sqrt{13}+...+\sqrt{2001}-\sqrt{2005}\right)\)
=> \(Q=-\frac{1}{4}.\left(1-\sqrt{2005}\right)\)
=> \(Q=\frac{\sqrt{2005}-1}{4}\)
\(=\sqrt{5}-1+\sqrt{9}-\sqrt{5}+...+\sqrt{2005}-\sqrt{2001}\)
\(=\sqrt{2005}-1\)
ở đây nhé :
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\(A=\frac{1}{\sqrt{2001}+\sqrt{2003}}+\frac{1}{\sqrt{2003}+\sqrt{2005}}+...+\frac{1}{\sqrt{2015}+\sqrt{2017}}\)
Ta có công thức:
\(\frac{1}{\sqrt{n}+\sqrt{n+1}}=\frac{\sqrt{n+1}-\sqrt{n}}{\left(\sqrt{n}+\sqrt{n+1}\right)\left(\sqrt{n+1}-\sqrt{n}\right)}=\sqrt{n+1}-\sqrt{n}\)
Áp dụng vào công thức ta có:
\(A=\frac{1}{\sqrt{2001}+\sqrt{2003}}+\frac{1}{\sqrt{2003}+\sqrt{2005}}+...+\frac{1}{\sqrt{2015}+\sqrt{2017}}\)
\(A=\sqrt{2003}-\sqrt{2001}+\sqrt{2005}-\sqrt{2003}+...+\sqrt{2017}-\sqrt{2015}\)
\(A=\sqrt{2017}-\sqrt{2001}\approx0,17848\)