\(\dfrac{1}{3}\) + \(\dfrac{1}{15}\) + \(\dfrac{1}{40}\) + ... + \(\dfrac{1}{101,103}\)
K
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Những câu hỏi liên quan
27 tháng 2 2022
a: \(=\left(9-\dfrac{13}{18}\right):\dfrac{325}{27}-\dfrac{17}{8}:\dfrac{51}{40}\)
\(=\dfrac{149}{18}\cdot\dfrac{27}{325}-\dfrac{17}{8}\cdot\dfrac{40}{51}\)
\(=\dfrac{447}{650}-\dfrac{5}{3}=-\dfrac{1909}{1950}\)
b: \(=\dfrac{48}{64}+\left(\dfrac{4}{5}-2-\dfrac{4}{15}\right):\dfrac{11}{3}\)
\(=\dfrac{3}{4}+\dfrac{-22}{15}\cdot\dfrac{3}{11}=\dfrac{3}{4}-\dfrac{2}{5}=\dfrac{15-8}{20}=\dfrac{7}{20}\)
8 tháng 3 2023
1, \(\dfrac{3}{4}.\left(\dfrac{2}{5}-\dfrac{1}{15}\right)+\dfrac{3}{4}=\dfrac{3}{4}.\left(\dfrac{2}{5}-\dfrac{1}{15}+1\right)\)
\(=\dfrac{3}{4}.\dfrac{6-1+15}{15}=\dfrac{3}{4}.\dfrac{20}{15}=\dfrac{3}{4}.\dfrac{4}{3}=1\)
2, \(\dfrac{4}{9}.\left(-\dfrac{13}{3}\right)+\dfrac{4}{3}.\dfrac{40}{9}=\dfrac{4}{9}.\left(-\dfrac{13}{3}\right)+\dfrac{4}{9}.\dfrac{40}{3}\)
\(=\dfrac{4}{9}.\left[\left(-\dfrac{13}{3}\right)+\dfrac{40}{3}\right]=\dfrac{4}{9}.9=4\)
3, \(\dfrac{4}{9}-\dfrac{2}{3}.\left(\dfrac{4}{5}+\dfrac{1}{2}\right)=\dfrac{2}{3}\left(\dfrac{2}{3}-\dfrac{4}{5}-\dfrac{1}{2}\right)\)
\(=\dfrac{2}{3}.\dfrac{20-24-15}{30}=\dfrac{2}{3}.\left(-\dfrac{19}{30}\right)=-\dfrac{19}{45}\)
TH
Thầy Hùng Olm
Manager
VIP
8 tháng 3 2023
1. \(\dfrac{3}{4}.\left(\dfrac{6}{15}-\dfrac{1}{15}\right)+\dfrac{3}{4}=\dfrac{3}{4}.\dfrac{1}{3}+\dfrac{3}{4}=\dfrac{1}{4}+\dfrac{3}{4}=1\)
NT
2
10 tháng 4 2022
\(=\dfrac{17}{8}:\dfrac{51}{40}-\left(15-13-\dfrac{1}{3}\right):\dfrac{25}{6}\)
\(=\dfrac{5}{3}-\dfrac{5}{3}\cdot\dfrac{6}{25}=\dfrac{5}{3}\cdot\dfrac{19}{25}=\dfrac{19}{15}\)
10 tháng 4 2022
\(2\dfrac{1}{8}:1\dfrac{11}{40}-\left(15-13\dfrac{1}{3}\right):4\dfrac{1}{6}\)
\(=\dfrac{17}{8}.\dfrac{40}{51}-\left(15-\dfrac{40}{3}\right).\dfrac{6}{25}\)
\(=\dfrac{17}{8}.\dfrac{40}{51}-\dfrac{5}{3}.\dfrac{6}{25}\)
\(=\dfrac{5}{3}-\dfrac{2}{5}\)
\(=\dfrac{19}{15}\)
GL
24 tháng 8 2023
a) \(\dfrac{4}{24}+\dfrac{7}{6}=\dfrac{4}{24}+\dfrac{28}{24}=\dfrac{4+28}{24}=\dfrac{32}{24}=\dfrac{4}{3}\)
b) \(\dfrac{10}{15}-\dfrac{1}{3}=\dfrac{10}{15}-\dfrac{5}{15}=\dfrac{10-5}{15}=\dfrac{5}{15}=\dfrac{1}{3}\)
c) \(\dfrac{21}{28}-\dfrac{1}{4}=\dfrac{21}{28}-\dfrac{7}{28}=\dfrac{21-7}{28}=\dfrac{14}{28}=\dfrac{1}{2}\)
d) \(\dfrac{35}{40}+\dfrac{5}{8}=\dfrac{35}{40}+\dfrac{25}{40}=\dfrac{35+25}{40}=\dfrac{60}{40}=\dfrac{3}{2}\)
24 tháng 8 2023
\(a)\dfrac{4}{24}=\dfrac{1}{6} \\ \dfrac{1}{6}+\dfrac{7}{6}\\ =\dfrac{8}{6}=\dfrac{4}{3}\\ b)\dfrac{10}{15}=\dfrac{2}{3}-\dfrac{1}{3}\\ =\dfrac{1}{3}\\ c)\dfrac{21}{28}=\dfrac{3}{4}\\ \dfrac{3}{4}-\dfrac{1}{4}\\ =\dfrac{2}{4}=\dfrac{1}{2}\\ d)\dfrac{35}{40}=\dfrac{7}{8}\\ \dfrac{7}{8}+\dfrac{5}{8}\\ =\dfrac{12}{8}=\dfrac{3}{2}\)
19 tháng 8 2021
a: Ta có: \(\dfrac{1}{4}:x=3\dfrac{4}{5}:40\dfrac{8}{15}\)
\(\Leftrightarrow x=\dfrac{1}{4}\cdot\dfrac{\dfrac{608}{15}}{3+\dfrac{4}{5}}\)
\(\Leftrightarrow x=\dfrac{152}{15}:\dfrac{19}{5}=\dfrac{8}{3}\)
b: Ta có: \(\left(x+1\right):\dfrac{5}{6}=\dfrac{20}{3}\)
\(\Leftrightarrow x+1=\dfrac{50}{9}\)
hay \(x=\dfrac{41}{9}\)
c: Ta có: \(\dfrac{7}{x-1}=\dfrac{x+1}{9}\)
\(\Leftrightarrow x^2-1=63\)
\(\Leftrightarrow x^2=64\)
hay \(x\in\left\{8;-8\right\}\)
NT
16 tháng 10 2022
c. \(\dfrac{7}{x-1}=\dfrac{x+1}{9}\)
\(7.9=\left(x-1\right).\left(x+1\right)\)
\(63=x^2-1\)
\(x^2=63+1\)
\(x^2=64\)
\(x^2=8^2\)
\(x=8\)
LV
2
27 tháng 8 2022
a: \(=\dfrac{2}{3}\left(\dfrac{3}{60\cdot63}+\dfrac{3}{63\cdot66}+...+\dfrac{3}{117\cdot120}\right)+\dfrac{2}{2006}\)
\(=\dfrac{2}{3}\left(\dfrac{1}{60}-\dfrac{1}{63}+...+\dfrac{1}{117}-\dfrac{1}{120}\right)+\dfrac{2}{2006}\)
\(=\dfrac{2}{3}\cdot\dfrac{1}{120}+\dfrac{1}{2003}=\dfrac{1}{180}+\dfrac{1}{2003}=\dfrac{2183}{180\cdot2003}\)
b: \(=\dfrac{5}{4}\left(\dfrac{4}{40\cdot44}+\dfrac{4}{44\cdot48}+...+\dfrac{4}{76\cdot80}\right)+\dfrac{5}{2006}\)
\(=\dfrac{5}{4}\left(\dfrac{1}{40}-\dfrac{1}{80}\right)+\dfrac{5}{2006}\)
\(=\dfrac{5}{4}\cdot\dfrac{1}{80}+\dfrac{5}{2006}=\dfrac{1}{64}+\dfrac{5}{2006}=\dfrac{1163}{64192}\)
c: \(=\dfrac{1}{3}\left(\dfrac{3}{2\cdot5}+\dfrac{3}{5\cdot8}+\dfrac{3}{8\cdot11}+\dfrac{3}{11\cdot14}+\dfrac{3}{14\cdot17}+\dfrac{3}{17\cdot20}\right)\)
\(=\dfrac{1}{3}\left(\dfrac{1}{2}-\dfrac{1}{20}\right)=\dfrac{1}{3}\cdot\dfrac{9}{20}=\dfrac{3}{20}\)
