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\(\dfrac{1}{3\cdot4}-\dfrac{1}{4\cdot5}-...-\dfrac{1}{9\cdot10}\)

\(=\dfrac{1}{3}-\dfrac{1}{4}-\left(\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{9}-\dfrac{1}{10}\right)\)

\(=\dfrac{1}{3}-\dfrac{1}{2}+\dfrac{1}{10}\)

\(=\dfrac{10}{30}-\dfrac{15}{30}+\dfrac{3}{30}\)

\(=\dfrac{-1}{15}\)

a) Ta có: \(\dfrac{-5}{7}\left(\dfrac{14}{5}-\dfrac{7}{10}\right):\left|-\dfrac{2}{3}\right|-\dfrac{3}{4}\left(\dfrac{8}{9}+\dfrac{16}{3}\right)+\dfrac{10}{3}\left(\dfrac{1}{3}+\dfrac{1}{5}\right)\)

\(=\dfrac{-5}{7}\cdot\dfrac{3}{2}\cdot\dfrac{21}{10}-\dfrac{3}{4}\cdot\dfrac{56}{3}+\dfrac{10}{3}\cdot\dfrac{8}{15}\)

\(=\dfrac{-9}{4}-14+\dfrac{16}{9}\)

\(=\dfrac{-1621}{126}\)

b) Ta có: \(\dfrac{17}{-26}\cdot\left(\dfrac{1}{6}-\dfrac{5}{3}\right):\dfrac{17}{13}-\dfrac{20}{3}\left(\dfrac{2}{5}-\dfrac{1}{4}\right)+\dfrac{2}{3}\left(\dfrac{6}{5}-\dfrac{9}{2}\right)\)

\(=\dfrac{-17}{26}\cdot\dfrac{13}{17}\cdot\dfrac{-3}{2}-\dfrac{20}{3}\cdot\dfrac{3}{20}+\dfrac{2}{3}\cdot\dfrac{-33}{10}\)

\(=\dfrac{3}{4}-1-\dfrac{11}{5}\)

\(=-\dfrac{49}{20}\)

\(A=\frac{9}{1}+\frac{8}{2}+\frac{7}{3}+\frac{6}{4}+\frac{5}{5}+\frac{4}{6}+\frac{3}{7}+\frac{2}{8}+\frac{1}{9}\)

\(=\left(9-1-1-...1\right)+\left(\frac{8}{2}+1\right)+\left(\frac{7}{3}+1\right)+...+\left(\frac{1}{9}+1\right)\)

\(=1+\frac{10}{2}+\frac{10}{3}+...+\frac{10}{9}=\frac{10}{2}+\frac{10}{3}+...+\frac{10}{9}+\frac{10}{10}\)

\(=10\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}+\frac{1}{9}\right)=10B\)

vậy A:B=10

3 tháng 11 2023

(1/2-1/3)+(1/3-1/4)+(1/4-1/5)+(1/5-1/6)+(1/6-1/7)+(1/7-1/8)+(1/8-1/9)+(1/9-1/10)

=1/2-1/3+1/3-1/4+1/4-1/5+1/5-1/6+1/6-1/7+1/7-1/8+1/8-1/9+1/9-1/10

=1/2-1/10

=2/5

5: \(=3-\dfrac{1}{4}+\dfrac{2}{3}-5+\dfrac{1}{3}+\dfrac{6}{5}-6+\dfrac{7}{4}-\dfrac{3}{2}\)

\(=3-5-6+\dfrac{-1}{4}+\dfrac{7}{4}+\dfrac{2}{3}+\dfrac{1}{3}+\dfrac{6}{5}-\dfrac{3}{2}\)

\(=-8+\dfrac{3}{2}+1+\dfrac{-3}{10}\)

\(=-7+\dfrac{15-3}{10}=-7+\dfrac{6}{5}=-\dfrac{29}{5}\)

6: \(=6-\dfrac{2}{3}+\dfrac{1}{2}-5-\dfrac{5}{3}+\dfrac{3}{2}-3+\dfrac{7}{3}-\dfrac{5}{2}\)

\(=6-5-3-\dfrac{2}{3}-\dfrac{5}{3}+\dfrac{7}{3}+\dfrac{1}{2}+\dfrac{3}{2}-\dfrac{5}{2}\)

\(=-2-\dfrac{1}{2}=-\dfrac{5}{2}\)

7: \(=\dfrac{5}{3}-\dfrac{3}{7}+9-2-\dfrac{5}{7}+\dfrac{2}{3}+\dfrac{8}{7}-\dfrac{4}{3}-10\)

\(=9-2-10+\dfrac{5}{3}+\dfrac{2}{3}-\dfrac{4}{3}+\dfrac{-3}{7}-\dfrac{5}{7}+\dfrac{8}{7}\)

=-3+1

=-2

8: \(=8-\dfrac{9}{4}+\dfrac{2}{7}+6+\dfrac{3}{7}-\dfrac{5}{4}-3-\dfrac{2}{4}+\dfrac{9}{7}\)

\(=8+6-3+\dfrac{2}{7}+\dfrac{3}{7}+\dfrac{9}{7}-1-\dfrac{2}{4}\)

\(=11+2-1-\dfrac{1}{2}\)

=11+1/2

=11,5

30 tháng 9 2020

\(\frac{1}{10\times9}-\frac{1}{9\times8}-\frac{1}{8\times7}-\frac{1}{7\times6}-\frac{1}{6\times5}-\frac{1}{5\times4}-\frac{1}{4\times3}-\frac{1}{3\times2}-\frac{1}{2\times1}\)

\(=\frac{1}{10\times9}-\left(\frac{1}{9\times8}+\frac{1}{8\times7}+\frac{1}{7\times6}+\frac{1}{6\times5}+\frac{1}{5\times4}+\frac{1}{4\times3}+\frac{1}{3\times2}+\frac{1}{2\times1}\right)\)

\(=\frac{1}{90}-\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}\right)\)

\(=\frac{1}{90}-\left(\frac{1}{1}-\frac{1}{9}\right)\)

\(=\frac{1}{90}-\frac{8}{9}\)

\(=-\frac{79}{90}\)

3 tháng 10 2020

cảm ơn bạ

28 tháng 9 2021

hổng có bík

28 tháng 9 2021

\(B=\frac{1}{3.4}-\frac{1}{4.5}-\frac{1}{5.6}-\frac{1}{6.7}-\frac{1}{7.8}-\frac{1}{8.9}-\frac{1}{9.10}\)

\(=\frac{1}{3}-\frac{1}{4}-\left(\frac{1}{4}-\frac{1}{5}\right)-\left(\frac{1}{5}-\frac{1}{6}\right)-\left(\frac{1}{6}-\frac{1}{7}\right)-\left(\frac{1}{7}-\frac{1}{8}\right)-\left(\frac{1}{8}-\frac{1}{9}\right)-\left(\frac{1}{9}-\frac{1}{10}\right)\)

\(=\frac{1}{3}-\frac{1}{4}-\frac{1}{4}+\frac{1}{10}\)

\(=\frac{-1}{15}\)

15 tháng 10 2021

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