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\(1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{19}-\dfrac{1}{20}\)
\(=1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{19}-\dfrac{1}{20}+\left(\dfrac{1}{2}-\dfrac{1}{2}\right)+\left(\dfrac{1}{4}-\dfrac{1}{4}\right)+...+\left(\dfrac{1}{20}-\dfrac{1}{20}\right)\)
\(=1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{20}-2\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{20}\right)\)
\(=1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{20}-\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{10}\right)\)
\(=\dfrac{1}{11}+\dfrac{1}{12}+...+\dfrac{1}{20}\) (đpcm)
Ta có :
\(\dfrac{1}{11}>\dfrac{1}{20}\\ \dfrac{1}{12}>\dfrac{1}{20}\\ ..........\\ \dfrac{1}{20}=\dfrac{1}{20}\)
\(\Rightarrow\dfrac{1}{11}+\dfrac{1}{12}+\dfrac{1}{13}+...+\dfrac{1}{20}>\dfrac{1}{20}+\dfrac{1}{20}+...+\dfrac{1}{20}\\ \Rightarrow S>\dfrac{10}{20}\\ \Rightarrow S>\dfrac{1}{2}\)
Ta có: \(\dfrac{1}{11}>\dfrac{1}{20}\)
\(\dfrac{1}{12}>\dfrac{1}{20}\)
\(\dfrac{1}{13}>\dfrac{1}{20}\)
\(\dfrac{1}{14}>\dfrac{1}{20}\)
\(\dfrac{1}{15}>\dfrac{1}{20}\)
\(\dfrac{1}{16}>\dfrac{1}{20}\)
\(\dfrac{1}{17}>\dfrac{1}{20}\)
\(\dfrac{1}{18}>\dfrac{1}{20}\)
\(\dfrac{1}{19}>\dfrac{1}{20}\)
\(\dfrac{1}{20}=\dfrac{1}{20}\)
=> \(\dfrac{1}{11}+\dfrac{1}{12}+...+\dfrac{1}{20}>\dfrac{1}{20}.10\)
hay S > \(\dfrac{1}{2}\)
Ta có :
\(\dfrac{1}{11}>\dfrac{1}{20}\) ( vì 1 > 0 , 0 < 11 < 20 )
\(\dfrac{1}{12}>\dfrac{1}{20}\) ( vì 1 > 0 , 0 < 12 < 20 )
...
\(\dfrac{1}{20}=\dfrac{1}{20}\)
\(\Rightarrow\dfrac{1}{11}+\dfrac{1}{12}+\dfrac{1}{13}+...+\dfrac{1}{20}>\dfrac{1}{20}+\dfrac{1}{20}+...+\dfrac{1}{20}\)( 10 số hạng )
\(\Rightarrow S>\dfrac{1}{20}.10\Rightarrow S>\dfrac{10}{20}\Rightarrow S>\dfrac{1}{2}\)
Vậy ...
ta thấy : \(\dfrac{1}{11},\dfrac{1}{12},\dfrac{1}{13},...\dfrac{1}{19}\)đều lớn hơn\(\dfrac{1}{20}\)
=>\(\dfrac{1}{11}+\dfrac{1}{12}+\dfrac{1}{13}+...+\dfrac{1}{20}>\dfrac{1}{20}+\dfrac{1}{20}+...+\dfrac{1}{20}\)(20 số hạng \(\dfrac{1}{20}\))
=>\(\dfrac{1}{11}+\dfrac{1}{12}+\dfrac{1}{13}+..+\dfrac{1}{20}>1\) mà 1 > \(\dfrac{1}{2}\) =>\(\dfrac{1}{11}+\dfrac{1}{12}+\dfrac{1}{13}+..+\dfrac{1}{20}>\dfrac{1}{2}\)
Gợi ý: Sử dụng tính chất phân phối của phép nhân đối với phép cộng để nhóm thừa số chung ra ngoài.
1) \(\dfrac{1}{2}+\dfrac{13}{19}-\dfrac{4}{9}+\dfrac{6}{19}+\dfrac{5}{18}\)
\(=\dfrac{1}{2}+\left(\dfrac{13}{19}+\dfrac{6}{19}\right)-\dfrac{4}{9}+\dfrac{5}{18}\)
\(=\dfrac{3}{2}-\dfrac{4}{9}+\dfrac{5}{18}\)
\(=\dfrac{19}{18}+\dfrac{5}{18}\)
\(=\dfrac{24}{18}\)
\(=\dfrac{4}{3}\)
2) \(\dfrac{-20}{23}+\dfrac{2}{3}-\dfrac{3}{23}+\dfrac{2}{5}+\dfrac{7}{15}\)
\(=\left(-\dfrac{20}{23}-\dfrac{3}{23}\right)+\dfrac{2}{3}+\dfrac{2}{5}+\dfrac{7}{15}\)
\(=-1+\dfrac{2}{3}+\dfrac{2}{5}+\dfrac{7}{15}\)
\(=-\dfrac{1}{3}+\dfrac{2}{5}+\dfrac{7}{15}\)
\(=\dfrac{1}{15}+\dfrac{7}{15}\)
\(=\dfrac{8}{15}\)
3) \(\dfrac{4}{3}+\dfrac{-11}{31}+\dfrac{3}{10}-\dfrac{20}{31}-\dfrac{2}{5}\)
\(=\left(\dfrac{-11}{31}-\dfrac{20}{31}\right)+\dfrac{4}{3}+\dfrac{3}{10}-\dfrac{2}{5}\)
\(=-1+\dfrac{4}{3}+\dfrac{3}{10}-\dfrac{2}{5}\)
\(=\dfrac{1}{3}+\dfrac{3}{10}-\dfrac{2}{5}\)
\(=\dfrac{1}{3}-\dfrac{1}{10}\)
\(=\dfrac{7}{30}\)
4) \(\dfrac{5}{7}.\dfrac{5}{11}+\dfrac{5}{7}.\dfrac{2}{11}-\dfrac{5}{7}.\dfrac{14}{11}\)
\(=\dfrac{5}{7}.\left(\dfrac{5}{11}+\dfrac{2}{11}-\dfrac{14}{11}\right)\)
\(=\dfrac{5}{7}.-\dfrac{7}{11}\)
\(=-\dfrac{35}{77}\)
\(=-\dfrac{5}{11}\)
Ta có:\(\dfrac{1}{13}+\dfrac{1}{14}+\dfrac{1}{15}+\dfrac{1}{16}>4\cdot\dfrac{1}{16}=\dfrac{1}{4}\)
\(\dfrac{1}{17}+\dfrac{1}{18}+\dfrac{1}{19}+\dfrac{1}{20}>4\cdot\dfrac{1}{20}=\dfrac{1}{5}\)
=>\(\dfrac{1}{13}+\dfrac{1}{14}+...+\dfrac{1}{20}>\dfrac{1}{4}+\dfrac{1}{5}=\dfrac{9}{20}\)
=>A>\(\dfrac{1}{12}+\dfrac{9}{20}\)
\(\dfrac{1}{12}>\dfrac{1}{20}\)
=>\(A>\dfrac{1}{20}+\dfrac{9}{20}=\dfrac{1}{2}\)
Vậy...
