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\(ĐK:\left\{{}\begin{matrix}x-2008\ge0\\2008-x\ge0\\x-2007>0\end{matrix}\right.\Leftrightarrow x=2008\)
Vậy PT có nghiệm \(x=2008\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 2:
\(P=\frac{1}{1+\sqrt{5}}+\frac{1}{\sqrt{5}+\sqrt{9}}+...+\frac{1}{\sqrt{2001}+\sqrt{2005}}\)
\(=\frac{1-\sqrt{5}}{\left(1+\sqrt{5}\right)\left(1-\sqrt{5}\right)}+\frac{\sqrt{5}-\sqrt{9}}{\left(\sqrt{5}+\sqrt{9}\right)\left(\sqrt{5}-\sqrt{9}\right)}+...+\frac{\sqrt{2001}-\sqrt{2005}}{\left(\sqrt{2001}+\sqrt{2005}\right)\left(\sqrt{2001}-\sqrt{2005}\right)}\)
\(=\frac{1-\sqrt{5}}{1-5}+\frac{\sqrt{5}-\sqrt{9}}{5-9}+...+\frac{\sqrt{2001}-\sqrt{2005}}{2001-2005}\)
\(=\frac{1-\sqrt{5}}{-4}+\frac{\sqrt{5}-\sqrt{9}}{-4}+..+\frac{\sqrt{2001}-\sqrt{2005}}{-4}\)
\(=\frac{1-\sqrt{5}+\sqrt{5}-\sqrt{9}+...+\sqrt{2001}-\sqrt{2005}}{-4}\)
\(=\frac{1-\sqrt{2005}}{-4}\)
\(=\frac{\sqrt{2005}-1}{4}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{1}{\sqrt{2009}-\sqrt{2008}}=\frac{\sqrt{2009}+\sqrt{2008}}{\left(\sqrt{2009}+\sqrt{2008}\right)\left(\sqrt{2009}-\sqrt{2008}\right)}=\frac{\sqrt{2009}+\sqrt{2008}}{2009-2008}=\sqrt{2009}+\sqrt{2008}\)
CMTT : \(\frac{1}{\sqrt{2008}-\sqrt{2007}}=\sqrt{2008}+\sqrt{2007}\)
Vì \(\sqrt{2009}+\sqrt{2008}>\sqrt{2008}+\sqrt{2007}\)
=> \(\frac{1}{\sqrt{2009}-\sqrt{2008}}\sqrt{2008}-\sqrt{2007}\)
S =2\(\sqrt{28}\)
P =1
X2 - 2\(\sqrt{28}\) X +1 =0
sao mình tính \(S=4\sqrt{502}\)