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12 tháng 5 2021
Ta có: \(C=\dfrac{\dfrac{2006}{2}+\dfrac{2006}{3}+\dfrac{2006}{4}+...+\dfrac{2006}{2007}}{\dfrac{2006}{1}+\dfrac{2005}{2}+\dfrac{2004}{3}+...+\dfrac{1}{2006}}\)
\(=\dfrac{\dfrac{2006}{2}+\dfrac{2006}{3}+\dfrac{2006}{4}+...+\dfrac{2006}{2007}}{1+\left(1+\dfrac{2005}{2}\right)+\left(1+\dfrac{2004}{3}\right)+...+\left(1+\dfrac{1}{2006}\right)}\)
\(=\dfrac{\dfrac{2006}{2}+\dfrac{2006}{3}+\dfrac{2006}{4}+...+\dfrac{2006}{2007}}{\dfrac{2007}{2007}+\dfrac{2007}{2}+\dfrac{2007}{3}+...+\dfrac{2007}{2006}}\)
\(=\dfrac{2006}{2007}\)
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MH
5
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NH
1
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18 tháng 12 2022
=>3n+3+5 chia hết cho n+1
=>\(n+1\in\left\{1;-1;5;-5\right\}\)
=>\(n\in\left\{0;-2;4;-6\right\}\)
\(\dfrac{1}{1\times2}+\dfrac{1}{2\times3}+...+\dfrac{1}{x\left(x+1\right)}=\dfrac{2008}{2009}\)
\(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{x}-\dfrac{1}{x+1}=\dfrac{2008}{2009}\)
\(1-\dfrac{1}{x+1}=\dfrac{2008}{2009}\)
\(\dfrac{x+1-1}{x+1}=\dfrac{2008}{2009}\)
\(\dfrac{x}{x+1}=\dfrac{2008}{2009}\)
\(\Leftrightarrow2008x+2008=2009x\)
\(\Leftrightarrow x=2008\)