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Ta có : \(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{x.\left(x+1\right)}=\frac{2008}{2009}\)
\(\Leftrightarrow\)\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{x.\left(x+1\right)}=\frac{2008}{2009}\)
\(\Leftrightarrow1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{2008}{2009}\)
\(\Leftrightarrow1-\frac{1}{x+1}=\frac{2008}{2009}\)
\(\Leftrightarrow\frac{1}{x+1}=\frac{1}{2009}\)
\(\Leftrightarrow x+1=2009\)
\(\Leftrightarrow x=2008\)
Vậy x = 2008
\(=>\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+......+\frac{1}{x.\left(x+1\right)}=\frac{2008}{2009}\)
\(=>1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+....+\frac{1}{x}-\frac{1}{x+1}=\frac{2008}{2009}\)
\(=>1-\frac{1}{x+1}=\frac{2008}{2009}\)
\(=>\frac{x}{x+1}=\frac{2008}{2009}=>x=2008\)
C=1.3.5.7...99
=>2.4.6...100.C=1.2.3...100
=>C = (1.2.3....100) / (2.4.6...100)= (1.2.3...50).(51.52...100) / [(2.1)(2.2).(2.3)...(2.50)]
C=(1.2.3...50).(51.52...100) /[2^50.(1.2.3...50)] =(51.52...100)/2^50 =51/2.52/2.53/2...100/2 =D
VAy C=D