\(\frac{8}{3}\) . \(\frac{2}{5}\)...">
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j) \(\frac{8}{3}.\frac{2}{5}.\frac{3}{8}.10.\frac{19}{92}=\left(\frac{8}{3}.\frac{3}{8}\right).\left(\frac{2}{5}.10\right).\frac{19}{92}=1.4.\frac{19}{92}\)

\(=\frac{19}{23}\)

k)\(\frac{-5}{7}.\frac{2}{11}.\frac{-13}{17}.\frac{19}{12}.\frac{17}{13}=\left(\frac{2}{11}.\frac{19}{12}\right).\left(\frac{-13}{17}.\frac{17}{13}\right).\frac{-5}{7}\)

\(=\frac{-19}{66}.\frac{-5}{7}=\frac{95}{462}\)

l)\(\frac{12}{19}.\frac{7}{15}.\frac{-13}{17}.\frac{19}{12}.\frac{17}{13}=\left(\frac{12}{19}.\frac{19}{12}\right).\left(\frac{-13}{17}.\frac{17}{13}\right).\frac{7}{15}\)

\(=\frac{-7}{15}\)

cậu tham khảo trên này ạ, nếu đúng cho mk 1 t.i.c.k ạ, thank nhiều

ta có;

b=8/3.2/5.3/8.10.19/92

b=16/15.3/8.10.19/92

b=2/5.10.19/92

b=4.19/92

b=19/23

c=-5/7.2/7+-5/7 . 9/14+1/5/7

c=-10/49+(-45)/98+1/5/5

c=131/98

25 tháng 7 2021

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

\(3\frac{14}{19}+\frac{13}{17}+\frac{35}{43}+6\frac{5}{19}+\frac{8}{43}\)

\(=\left(3\frac{14}{19}+6\frac{5}{19}\right)+\left(\frac{35}{43}+\frac{8}{43}\right)+\frac{13}{17}\)

\(=10+1+\frac{13}{17}=11+\frac{13}{17}=11\frac{13}{17}\)

1 tháng 9 2020

bạn chăm giải thật đó ~!

25 tháng 4 2019

A=19/20

25 tháng 4 2019

\(A=\frac{3}{2}-\frac{5}{6}+\frac{7}{12}-\frac{9}{20}+\frac{11}{30}-\frac{13}{42}+\frac{15}{56}-\frac{17}{72}\)

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

\(=1+\frac{1}{2}-\frac{1}{2}-\frac{1}{3}+\frac{1}{3}+\frac{1}{4}-\frac{1}{4}-\frac{1}{5}+\frac{1}{5}+\frac{1}{6}-...-\frac{1}{9}\)

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

\(=\frac{8}{9}\)

\(3\frac{14}{19}+\frac{13}{17}+\frac{35}{43}+6\frac{5}{19}+\frac{8}{43}\)

\(=\left(3\frac{14}{19}+6\frac{5}{19}\right)+\left(\frac{35}{43}+\frac{8}{43}\right)+\frac{13}{17}\)

\(=10+1+\frac{13}{17}=11+\frac{13}{17}=11\frac{13}{17}\)

\(\frac{-5}{7}.\frac{2}{11}+\frac{-5}{7}.\frac{9}{11}+1\frac{5}{7}\)

\(=\frac{-5}{7}.\frac{2}{11}+\frac{-5}{7}.\frac{9}{11}+1+\frac{-5}{7}.\left(-1\right)\)

\(=\frac{-5}{7}\left(\frac{2}{11}+\frac{9}{11}-1\right)+1\)

\(=\frac{-5}{7}.0+1==0+1=1\)