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A=\(\frac{1}{15}\left(\frac{1}{2}-\frac{1}{17}+\frac{1}{3}-\frac{1}{18}+...+\frac{1}{1990}-\frac{1}{2005}\right)\)
=\(\frac{1}{15}\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{1990}-\frac{1}{17}-\frac{1}{18}-...-\frac{1}{2005}\right)\)
=\(\frac{1}{15}\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{16}+\frac{1}{17}+\frac{1}{18}+...+\frac{1}{1990}-\frac{1}{17}-\frac{1}{18}-...-\frac{1}{1990}-...-\frac{1}{2005}\right)\)
=\(\frac{1}{15}\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{16}-\frac{1}{1991}-\frac{1}{1992}-...-\frac{1}{2005}\right)\)
B=\(\frac{1}{1989}\left(\frac{1}{2}-\frac{1}{1991}+\frac{1}{3}-\frac{1}{1992}+...+\frac{1}{16}-\frac{1}{2005}\right)\)
=\(\frac{1}{1989}\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{16}-\frac{1}{1991}-\frac{1}{1992}-...-\frac{1}{2005}\right)\)
2 dấu ngoặc của A và B là như nhau
Vậy A/B=1/15:1/1989=1/15.1989=663/5 ( đpcm, tức là điều phải chứng minh)
a) $A=\dfrac{1}{1.2}+\dfrac{1}{3.4}+...+\dfrac{1}{99.100}$
$=>A=1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}$
$=>A=(1+\dfrac{1}{3}+...+\dfrac{1}{99})-(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{100})$
$=>A=(1+\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{99}+\dfrac{1}{100})-(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{100}.2)$
$=>A=(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{100})-(1+\dfrac{1}{2}+...+\dfrac{1}{50})$
$=>A=\dfrac{1}{51}+\dfrac{1}{52}+...+\dfrac{1}{100}$
b) Ta có : $A=1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}$
$=>A=(1-\dfrac{1}{2}+\dfrac{1}{3})-(\dfrac{1}{4}-\dfrac{1}{5})-...-(\dfrac{1}{98}-\dfrac{1}{99})-\dfrac{1}{100}$
$=>A<1-\dfrac{1}{2}+\dfrac{1}{3}=\dfrac{5}{6}$
