Tìm X
$(\frac{1}{1.2}+ \frac{1}{3.4} +...+\frac{1}{99.100})X= \frac{2012}{51}+\frac{2012}{52}+...+\frac{2012}{99}+\frac{2012}{100}$
Ta thấy
$\frac{1}{1.2} = 1- \frac{1}{2}$ ; $\frac{1}{3.4} = \frac{1}{3}-\frac{1}{4}$;......;$\frac{1}{99.100}=\frac{1}{99}.\frac{1}{100}$
=> $(1-\frac{1}{2} + \frac{1}{3}-\frac{1}{4}+..........+\frac{1}{99}-\frac{1}{100}).X=
\frac{2012}{51}+\frac{2012}{52}+...+\frac{2012}{99}+\frac{2012}{100}$
=>$(1+ \frac{1}{3} + \frac{1}{5}+.....+\frac{1}{99}) - \frac{1}{2}-\frac{1}{4}-...-\frac{1}{100})X= \frac{2012}{51}+\frac{2012}{52}+...+\frac{2012}{99}+\frac{2012}{100}$
=> $(1+ \frac{1}{3} + \frac{1}{5}+.....+\frac{1}{99}) - (\frac{1}{2}+\frac{1}{4}+...+\frac{1}{100})X= \frac{2012}{51}+\frac{2012}{52}+...+\frac{2012}{99}+\frac{2012}{100}$
Ta tính ở : $(1+ \frac{1}{3} + \frac{1}{5}+.....+\frac{1}{99}) - (\frac{1}{2}+\frac{1}{4}+...+\frac{1}{100})$
Ta cộng $\frac{1}{2}+\frac{1}{4}+...+\frac{1}{100})$ vào số bị trừ và số trừ.
Ta có: $(1+ \frac{1}{3} + \frac{1}{5}+.....+\frac{1}{99}) - (\frac{1}{2}+\frac{1}{4}+...+\frac{1}{100})$
=$(1+ \frac{1}{3} + \frac{1}{5}+.....+\frac{1}{99}+\frac{1}{2}+\frac{1}{4}+...+\frac{1}{100}) - 2(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{100})$
=$(1+ \frac{1}{2}+\frac{1}{3}+.......+\frac{1}{100})- 2(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{100})$
=$(1+ \frac{1}{2}+\frac{1}{3}+.......+\frac{1}{100}(- (1+\frac{1}{2}+...+\frac{1}{50})$
=$\frac{1}{51}+\frac{1}{52}+....+\frac{1}{100}$
=> $(\frac{1}{51}+\frac{1}{52}+....+\frac{1}{100})X= \frac{2012}{51}+\frac{2012}{52}+...+\frac{2012}{99}+\frac{2012}{100}$
=> $(\frac{1}{51}+\frac{1}{52}+....+\frac{1}{100})X=2012.( \frac{1}{51}+\frac{1}{52}+...+\frac{1}{99}+\frac{1}{100})$
=> $X=2012$
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