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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 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 > \(\dfrac{1}{2}\) =>\(\dfrac{1}{11}+\dfrac{1}{12}+\dfrac{1}{13}+..+\dfrac{1}{20}>\dfrac{1}{2}\)
Ta có \(\dfrac{6}{15}>\dfrac{6}{16}>...>\dfrac{6}{19}\) nên \(S< \dfrac{6}{15}.5=2\).
Lại có \(S>\dfrac{6}{19}.5>1\) nên \(1< S< 2\)
a, \(\dfrac{14}{13}-\dfrac{1}{13}-\dfrac{19}{20}=1-\dfrac{19}{20}=\dfrac{1}{20}\)
b, \(-\dfrac{24}{17}+\dfrac{7}{17}+\dfrac{1}{16}=\dfrac{-17}{17}+\dfrac{1}{16}=-1+\dfrac{1}{16}=-\dfrac{15}{16}\)
1) âm năm phần 12
2) âm mười bảy phần 9
3) -1
Đây là đáp án còn làm bài từ làm nhé
Ta có: A\(=\dfrac{1}{9}.\dfrac{1}{10}+\dfrac{1}{10}.\dfrac{1}{11}+\dfrac{1}{11}.\dfrac{1}{12}+\dfrac{1}{12}.\dfrac{1}{13}+\dfrac{1}{13}.\dfrac{1}{14}+\dfrac{1}{14}.\dfrac{1}{15}\)
\(=\dfrac{1}{9}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{12}+\dfrac{1}{12}-\dfrac{1}{13}+\dfrac{1}{13}-\dfrac{1}{14}+\dfrac{1}{14}-\dfrac{1}{15}\)
\(=\dfrac{1}{9}-\dfrac{1}{15}=\dfrac{2}{45}\)
\(A=\dfrac{1}{9}.\dfrac{1}{10}+\dfrac{1}{10}.\dfrac{1}{11}+\dfrac{1}{11}.\dfrac{1}{12}+\dfrac{1}{12}.\dfrac{1}{13}+\dfrac{1}{13}.\dfrac{1}{14}+\dfrac{1}{14}.\dfrac{1}{15}\)
\(=\dfrac{1}{9}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{12}+\dfrac{1}{12}-\dfrac{1}{13}+\dfrac{1}{13}-\dfrac{1}{14}+\dfrac{1}{14}-\dfrac{1}{15}\)
\(=\dfrac{1}{9}-\dfrac{1}{15}\)
\(=\dfrac{2}{45}\)
-Chúc bạn học tốt-
. Ta có :
\(\dfrac{1}{11}>\dfrac{1}{20}\)
\(\dfrac{1}{12}>\dfrac{1}{20}\)
.................
\(\dfrac{1}{19}>\dfrac{1}{20}\)
\(\dfrac{1}{20}=\dfrac{1}{20}\)
\(\Leftrightarrow\dfrac{1}{11}+\dfrac{1}{12}+......+\dfrac{1}{20}>\dfrac{1}{20}+\dfrac{1}{20}+.....+\dfrac{1}{20}\)
\(\Leftrightarrow S>\dfrac{1}{20}.10\)
\(\Leftrightarrow S>\dfrac{1}{2}\)
2. \(\dfrac{x}{12}=\dfrac{-1}{24}-\dfrac{1}{8}\)
\(\Leftrightarrow\dfrac{x}{12}=-\dfrac{1}{6}\)
\(\Leftrightarrow6x=-12\)
\(\Leftrightarrow x=-2\)
Vậy ...
3. \(\dfrac{2}{5.7}+\dfrac{2}{7.9}+........+\dfrac{2}{19.21}\)
\(=\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+......+\dfrac{1}{19}-\dfrac{1}{21}\)
\(=\dfrac{1}{5}-\dfrac{1}{21}\)
\(=\dfrac{16}{105}\)
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: \(\dfrac{1}{11}>\dfrac{1}{20}\)
\(\dfrac{1}{12}>\dfrac{1}{20}\)
\(\dfrac{1}{13}>\dfrac{1}{20}\)
\(\dfrac{1}{14}>\dfrac{1}{20}\)
\(\dfrac{1}{15}>\dfrac{1}{20}\)
\(\dfrac{1}{16}>\dfrac{1}{20}\)
\(\dfrac{1}{17}>\dfrac{1}{20}\)
\(\dfrac{1}{18}>\dfrac{1}{20}\)
\(\dfrac{1}{19}>\dfrac{1}{20}\)
\(\dfrac{1}{20}=\dfrac{1}{20}\)
=> \(\dfrac{1}{11}+\dfrac{1}{12}+...+\dfrac{1}{20}>\dfrac{1}{20}.10\)
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 ...