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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) \(\dfrac{5}{17}+\dfrac{7}{17}+\dfrac{13}{17}=\dfrac{5}{17}+\left(\dfrac{7}{17}+\dfrac{13}{17}\right)=\dfrac{5}{17}+\dfrac{20}{17}=\dfrac{25}{17}\)
a) \(\dfrac{3}{5}\times\dfrac{17}{21}+\dfrac{2}{5}\times\dfrac{17}{21}\)
\(=\dfrac{17}{21}\times\left(\dfrac{3}{5}+\dfrac{2}{5}\right)\)
\(=\dfrac{17}{21}\times1\)
\(=\dfrac{17}{21}\)
b) \(\dfrac{11}{19}\times\dfrac{2}{7}+\dfrac{5}{7}\times\dfrac{11}{19}\)
\(=\dfrac{11}{19}\times\left(\dfrac{2}{7}+\dfrac{5}{7}\right)\)
\(=\dfrac{11}{19}\times1\)
\(=\dfrac{11}{19}\)
a: =17/21(3/5+2/5)
=17/21*5/5
=17/21
b: =11/19(2/7+5/7)
=11/19*1
=11/19
\(B=\dfrac{1+\dfrac{1}{7}+\dfrac{1}{7^2}-\dfrac{1}{7^3}}{4+\dfrac{4}{7}+\dfrac{4}{7^2}-\dfrac{4}{7^3}}\cdot\dfrac{858585}{313131}\cdot\left(-1\dfrac{14}{17}\right)\)
\(=\dfrac{1}{4}\cdot\dfrac{85}{31}\cdot\dfrac{-31}{17}\)
\(=\dfrac{-5}{4}\)
A = \(\dfrac{3}{2}\) - \(\dfrac{5}{6}\) + \(\dfrac{7}{12}\) - \(\dfrac{9}{20}\) + \(\dfrac{11}{30}\) - \(\dfrac{13}{42}\) + \(\dfrac{15}{56}\) - \(\dfrac{17}{72}\)
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 = \(\dfrac{8}{9}\)
\(A=\left(1+\dfrac{1}{2}\right)-\left(\dfrac{1}{2}+\dfrac{1}{3}\right)+\left(\dfrac{1}{3}+\dfrac{1}{4}\right)-\left(\dfrac{1}{4}+\dfrac{1}{5}\right)+\left(\dfrac{1}{5}+\dfrac{1}{6}\right)-\left(\dfrac{1}{6}+\dfrac{1}{7}\right)+\left(\dfrac{1}{7}+\dfrac{1}{8}\right)-\left(\dfrac{1}{8}+\dfrac{1}{9}\right)\)
\(A=1+\dfrac{1}{2}-\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}+\dfrac{1}{4}-\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}+\dfrac{1}{6}-\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}+\dfrac{1}{8}-\dfrac{1}{8}-\dfrac{1}{9}\)
\(A=1+\dfrac{1}{9}=\dfrac{10}{9}\)
\(A=\dfrac{10}{7.12}+\dfrac{10}{12.17}+\dfrac{10}{17.22}+...+\dfrac{10}{502.507}\) (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: \(\dfrac{135}{17}+\dfrac{85}{17}+\dfrac{65}{17}+\dfrac{15}{17}\)
\(=\left(\dfrac{135}{17}+\dfrac{65}{17}\right)+\left(\dfrac{85}{17}+\dfrac{15}{17}\right)\)
\(=\dfrac{200}{17}+\dfrac{100}{17}=\dfrac{300}{17}\)
b: Sửa đề: \(\dfrac{446}{200}+\dfrac{32}{200}+\dfrac{554}{200}+\dfrac{968}{200}\)
\(=\dfrac{446+554}{200}+\dfrac{32+968}{200}\)
\(=\dfrac{1000}{200}+\dfrac{1000}{200}\)
=5+5
=10
a)=(135/17 + 65/17) + (85/17+ 15/17)
=200/17 + 100/17
= 300/17
b)= (446/200 + 554/200) + (32/20 + 968/200)
= 1000/200 + (320/200 + 968/200)
= 1000/200 + 1288/200
= 3288/200 = 411/25