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\(\frac{4}{3\cdot7}+\frac{5}{7\cdot12}+\frac{1}{12\cdot13}+\frac{7}{13\cdot20}+\frac{3}{20\cdot23}\)
\(\frac{4}{3.7}+\frac{5}{7.12}+\frac{1}{12.13}+\frac{7}{13.20}+\frac{3}{20.23}=\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{12}+...+\frac{1}{20}-\frac{1}{23}=\frac{1}{3}-\frac{1}{23}=\frac{20}{69}\)
A=4/1.31+6/7.41+9/9.41+ 7/10.57
=20/35.31+30/35.41+45/45.41+35/50.57
=5(4/35.31+6/35.41+9/45.41+7/50.57)
=5(1/31-1/35+1/35-1/41+1/41-1/45+1/45-1/50+1/50-1/57)
=5(1/31-1/57)
B thì làm tương tự nhưng nhân với 2=> B=2(1/31-1/57)
=> A/B=5/2
a, 3 \(\frac{14}{19}\)+ \(\frac{13}{17}\)+ \(\frac{35}{43}\)+ 6\(\frac{5}{19}\)+ \(\frac{8}{43}\)= \(\left(3\frac{14}{19}+6\frac{5}{19}\right)+\left(\frac{35}{43}+\frac{8}{43}\right)+\frac{13}{17}=\)\(9+1+\frac{13}{17}=8+\frac{13}{17}=8\frac{13}{17}\)
b, \(\frac{-5}{7}.\frac{2}{11}+\frac{-5}{7}.\frac{9}{11}+1\frac{5}{7}\)\(=\frac{-5}{7}\left(\frac{2}{11}+\frac{9}{11}\right)+1\frac{5}{7}\)\(=\frac{-5}{7}.1+1\frac{5}{7}\)\(=\frac{-5}{7}+\frac{12}{7}=\frac{7}{7}=1\)
Chúc bn học tốt
\(3\frac{14}{19}+\frac{13}{17}+\frac{35}{43}+6\frac{5}{19}+\frac{8}{43}\)
\(=\left(3\frac{14}{19}+6\frac{5}{19}\right)+\left(\frac{35}{43}+\frac{8}{43}\right)+\frac{13}{17}\)
\(=10+1+\frac{13}{17}=11+\frac{13}{17}=11\frac{13}{17}\)
\(M=\frac{32}{323}\) \(N=\frac{86}{589}\) \(\frac{M}{N}=\frac{496}{731}\)
\(3\frac{14}{19}+\frac{13}{17}+\frac{35}{43}+6\frac{5}{19}+\frac{8}{43}\)
\(=\left(3\frac{14}{19}+6\frac{5}{19}\right)+\left(\frac{35}{43}+\frac{8}{43}\right)+\frac{13}{17}\)
\(=10+1+\frac{13}{17}=11+\frac{13}{17}=11\frac{13}{17}\)
\(\frac{-5}{7}.\frac{2}{11}+\frac{-5}{7}.\frac{9}{11}+1\frac{5}{7}\)
\(=\frac{-5}{7}.\frac{2}{11}+\frac{-5}{7}.\frac{9}{11}+1+\frac{-5}{7}.\left(-1\right)\)
\(=\frac{-5}{7}\left(\frac{2}{11}+\frac{9}{11}-1\right)+1\)
\(=\frac{-5}{7}.0+1==0+1=1\)
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Sửa lại đề bài \(S=\frac{7}{3.13}+\frac{7}{13.23}+\frac{7}{23.33}+\frac{7}{33.43}+\cdots+\frac{7}{53.63}\)
Theo đề bài ta có phân số quy luật sau
\(\frac{7}{\left(10n-7\right)\left(10n+3\right)}\left(n=1\rarr6\right)\)
\(=\frac{7}{10}\left(\frac{1}{10n-7}-\frac{1}{10n+3}\right)\)
\(\rArr S=\frac{7}{10}\left(\frac13-\frac{1}{13}+\frac{1}{13}-\frac{1}{23}+\frac{1}{23}-\frac{1}{33}+\frac{1}{33}-\frac{1}{43}+\frac{1}{43}-\frac{1}{53}+\frac{1}{53}-\frac{1}{63}\right)\) \(\rArr S=\frac{7}{10}\left(\frac13-\frac{1}{63}\right)=\frac{7}{10}.\frac{20}{63}=\frac29\)
Vậy \(S=\frac29\)