Tính bằng cách thuận tiện nhất.
a) \(\dfrac{16}{15}+\dfrac{7}{15}+\dfrac{4}{15}\) b) \(\dfrac{5}{17}+\dfrac{7}{17}+\dfrac{13}{17}\)
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a: =4/5+1/5+2/3+1/3=1+1=2
b: =17/12+7/12+29/7-8/7=3+2=5
c: =3/5+2/5+16/7-1/7-1/7
=1+2=3
d: =2/5+3/5+2/3+1/3+7/4+1/4
=1+1+2
=4
a: \(=\dfrac{8}{9}\cdot\dfrac{9}{4}\cdot\dfrac{12}{19}\cdot\dfrac{19}{24}=\dfrac{1}{2}\cdot2=1\)
b: \(=\dfrac{5}{16}\cdot\dfrac{17}{15}\cdot\dfrac{8}{17}=\dfrac{5}{16}\cdot\dfrac{8}{15}=\dfrac{40}{240}=\dfrac{1}{6}\)
c: \(=\dfrac{4}{13}\left(\dfrac{2}{7}+\dfrac{5}{7}\right)-\dfrac{3}{26}=\dfrac{4}{13}-\dfrac{3}{26}=\dfrac{5}{26}\)
c: \(=\dfrac{3}{4}\left(\dfrac{6}{11}+\dfrac{5}{11}\right)-\dfrac{1}{5}=\dfrac{3}{4}-\dfrac{1}{5}=\dfrac{11}{20}\)
A = \(\dfrac{3}{2}\) - \(\dfrac{5}{6}\) + \(\dfrac{7}{12}\) - \(\dfrac{9}{20}\) + \(\dfrac{11}{30}\) - \(\dfrac{13}{42}\) + \(\dfrac{15}{56}\) - \(\dfrac{17}{72}\)
A = (1 + \(\dfrac{1}{2}\)) - (\(\dfrac{1}{2}\) + \(\dfrac{1}{3}\)) + (\(\dfrac{1}{3}\) + \(\dfrac{1}{4}\)) - (\(\dfrac{1}{4}\) + \(\dfrac{1}{5}\)) + (\(\dfrac{1}{5}\) + \(\dfrac{1}{6}\)) - (\(\dfrac{1}{6}\) + \(\dfrac{1}{7}\)) + (\(\dfrac{1}{7}\) + \(\dfrac{1}{8}\)) - (\(\dfrac{1}{8}\) + \(\dfrac{1}{9}\))
A = 1 + \(\dfrac{1}{2}\) - \(\dfrac{1}{2}\) - \(\dfrac{1}{3}\) + \(\dfrac{1}{3}\) + \(\dfrac{1}{4}\) - \(\dfrac{1}{4}\) - \(\dfrac{1}{5}\) + \(\dfrac{1}{5}\) + \(\dfrac{1}{6}\) - \(\dfrac{1}{6}\) - \(\dfrac{1}{7}\) + \(\dfrac{1}{7}\) + \(\dfrac{1}{8}\) - \(\dfrac{1}{8}\) - \(\dfrac{1}{9}\)
A = 1 - \(\dfrac{1}{9}\)
A = \(\dfrac{8}{9}\)
\(A=\left(1+\dfrac{1}{2}\right)-\left(\dfrac{1}{2}+\dfrac{1}{3}\right)+\left(\dfrac{1}{3}+\dfrac{1}{4}\right)-\left(\dfrac{1}{4}+\dfrac{1}{5}\right)+\left(\dfrac{1}{5}+\dfrac{1}{6}\right)-\left(\dfrac{1}{6}+\dfrac{1}{7}\right)+\left(\dfrac{1}{7}+\dfrac{1}{8}\right)-\left(\dfrac{1}{8}+\dfrac{1}{9}\right)\)
\(A=1+\dfrac{1}{2}-\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}+\dfrac{1}{4}-\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}+\dfrac{1}{6}-\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}+\dfrac{1}{8}-\dfrac{1}{8}-\dfrac{1}{9}\)
\(A=1+\dfrac{1}{9}=\dfrac{10}{9}\)
\(a,\dfrac{5}{13}\times\dfrac{4}{15}\times13=\dfrac{5\times4\times13}{13\times5\times3}=\dfrac{4}{3}\\ b,\left(\dfrac{3}{7}+\dfrac{5}{2}\right)\times\dfrac{7}{5}=\dfrac{3}{7}\times\dfrac{7}{5}+\dfrac{5}{2}\times\dfrac{7}{5}=\dfrac{3}{5}+\dfrac{7}{2}=\dfrac{6}{10}+\dfrac{35}{10}=\dfrac{41}{10}\\ c,\dfrac{1}{5}\times\dfrac{11}{18}+\dfrac{11}{18}\times\dfrac{3}{5}=\dfrac{11}{18}\times\left(\dfrac{1}{5}+\dfrac{3}{5}\right)=\dfrac{11}{18}\times\dfrac{4}{5}=\dfrac{22}{45}\)
\(A=\dfrac{10}{7.12}+\dfrac{10}{12.17}+\dfrac{10}{17.22}+...+\dfrac{10}{502.507}\) (sửa 502+507 thành 503.507)
\(\Rightarrow A=10\left(\dfrac{1}{7.12}+\dfrac{1}{12.17}+\dfrac{1}{17.22}+...+\dfrac{1}{502.507}\right)\)
\(\Rightarrow A=10.\dfrac{1}{5}\left(\dfrac{1}{7}-\dfrac{1}{12}+\dfrac{1}{12}-\dfrac{1}{17}+\dfrac{1}{17}-\dfrac{1}{22}+...+\dfrac{1}{502}-\dfrac{1}{507}\right)\)
\(\Rightarrow A=2.\left(\dfrac{1}{7}-\dfrac{1}{507}\right)=2.\left(\dfrac{500}{3549}\right)=\dfrac{1000}{3549}\)
\(B=\dfrac{4}{8.13}+\dfrac{4}{13.18}+\dfrac{4}{18.23}+...+\dfrac{4}{253.258}\)
\(\Rightarrow B=4\left(\dfrac{1}{8.13}+\dfrac{1}{13.18}+\dfrac{1}{18.23}+...+\dfrac{1}{253.258}\right)\)
\(\Rightarrow B=4.\dfrac{1}{5}\left(\dfrac{1}{8}-\dfrac{1}{13}+\dfrac{1}{13}-\dfrac{1}{18}+\dfrac{1}{18}-\dfrac{1}{23}+...+\dfrac{1}{253}-\dfrac{1}{258}\right)\)
\(\Rightarrow B=\dfrac{4}{5}\left(\dfrac{1}{8}-\dfrac{1}{258}\right)=\dfrac{4}{5}\left(\dfrac{129}{1032}-\dfrac{8}{1032}\right)=\dfrac{4}{5}.\dfrac{121}{1032}=\dfrac{121}{1290}\)
a: \(=\left(\dfrac{3}{17}+\dfrac{14}{17}\right)+\left(\dfrac{-5}{13}-\dfrac{8}{13}\right)+\left(\dfrac{-18}{35}-\dfrac{17}{35}\right)\)
=1-1-1
=-1
b: \(=\dfrac{-3}{8}\left(\dfrac{1}{6}+\dfrac{5}{6}\right)+\dfrac{-5}{8}=\dfrac{-3}{8}-\dfrac{5}{8}=-1\)
c: \(=\dfrac{4}{4}\cdot\dfrac{5}{15}\cdot\dfrac{11}{11}=\dfrac{1}{3}\)
a)\(=\left(-\dfrac{5}{13}+\dfrac{-8}{13}\right)+\left(-\dfrac{18}{35}-\dfrac{17}{35}\right)+\left(\dfrac{3}{14}+\dfrac{14}{17}\right)=-1-1+1=-1\)
b)\(=\dfrac{-3}{8}.\left(\dfrac{1}{6}+\dfrac{5}{6}\right)-\dfrac{10}{16}=-\dfrac{3}{8}.1-\dfrac{10}{16}=-\dfrac{6}{16}-\dfrac{10}{16}=-\dfrac{16}{16}=-1\)
c)\(\dfrac{-4.5.11}{11.5.3.-4}=\dfrac{1}{3}\)
\(B=\dfrac{1+\dfrac{1}{7}+\dfrac{1}{7^2}-\dfrac{1}{7^3}}{4+\dfrac{4}{7}+\dfrac{4}{7^2}-\dfrac{4}{7^3}}\cdot\dfrac{858585}{313131}\cdot\left(-1\dfrac{14}{17}\right)\)
\(=\dfrac{1}{4}\cdot\dfrac{85}{31}\cdot\dfrac{-31}{17}\)
\(=\dfrac{-5}{4}\)
a) (689 + 875) + 125
= 689 + (875 + 125)
= 689 + 1000 = 1689.
581 + (878 + 419)
= (581 + 419) = 878
= 1000 + 878 = 1878.
b)
c) 5,87 + 28,69 + 4,13
= (5,87 + 4,13) + 28,69
= 10 + 28,69
= 38,69.
83,75 + 46,98 + 6,25
= (83,75 + 6,25) + 46,98
= 90 + 46,98
= 136,98.
a) \(\dfrac{16}{15}+\dfrac{7}{15}+\dfrac{4}{15}=\left(\dfrac{16}{15}+\dfrac{4}{15}\right)+\dfrac{7}{15}=\dfrac{20}{15}+\dfrac{7}{15}=\dfrac{27}{15}\)
b) \(\dfrac{5}{17}+\dfrac{7}{17}+\dfrac{13}{17}=\dfrac{5}{17}+\left(\dfrac{7}{17}+\dfrac{13}{17}\right)=\dfrac{5}{17}+\dfrac{20}{17}=\dfrac{25}{17}\)