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Tính giá trị biểu thức: 

B = 1\(\frac{6}{41}\). (\(\frac{12+\frac{12}{19}+\frac{12}{37}-\frac{12}{53}}{3+\frac{3}{19}+\frac{3}{37}-\frac{3}{53}}\):\(\frac{4+\frac{4}{19}+\frac{4}{37}-\frac{4}{53}}{5+\frac{5}{19}+\frac{5}{37}-\frac{5}{53}}\)) . \(\frac{124242423}{237373735}\)

Nhật Hạ
23 tháng 2 2020 lúc 16:45

\(B=1\frac{6}{41}.\left(\frac{12+\frac{12}{19}+\frac{12}{37}-\frac{12}{53}}{3+\frac{3}{19}+\frac{3}{37}-\frac{3}{53}}\right):\left(\frac{4+\frac{4}{19}+\frac{4}{37}-\frac{4}{53}}{5+\frac{5}{19}+\frac{5}{37}-\frac{5}{53}}\right).\frac{124242423}{237373735}\)

\(B=1\frac{6}{41}.\left[\frac{12\left(\frac{1}{19}+\frac{1}{37}-\frac{1}{53}\right)}{3\left(\frac{1}{19}+\frac{1}{37}-\frac{1}{53}\right)}\right]:\left[\frac{4\left(\frac{1}{19}+\frac{1}{37}-\frac{1}{53}\right)}{5\left(\frac{1}{19}+\frac{1}{37}-\frac{1}{53}\right)}\right].\frac{124242423}{237373735}\)

\(B=1\frac{6}{41}\left(\frac{12}{3}.\frac{5}{4}\right).\frac{124242423}{237373735}\)

\(B=1\frac{6}{41}.5.\frac{123}{235}\)

\(B=\frac{47.5.123}{41.235}=\frac{47.5.41.3}{41.5.47}=3\)

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╰Nguyễn Trí Nghĩa (team...
23 tháng 2 2020 lúc 17:15

B=\(1\frac{6}{41}.\left(\frac{12+\frac{12}{19}+\frac{12}{37}-\frac{12}{53}}{3+\frac{3}{19}+\frac{3}{37}-\frac{3}{53}}:\frac{4+\frac{4}{19}+\frac{4}{37}-\frac{4}{53}}{5+\frac{5}{19}+\frac{5}{37}-\frac{5}{53}}\right).\frac{124242423}{237373735}\)

B=\(\frac{47}{41}.\left(\frac{12.\left(1+\frac{1}{19}+\frac{1}{37}-\frac{1}{53}\right)}{3.\left(1+\frac{1}{19}+\frac{1}{37}-\frac{1}{53}\right)}:\frac{4.\left(1+\frac{1}{19}+\frac{1}{37}-\frac{1}{53}\right)}{5.\left(1+\frac{1}{19}+\frac{1}{37}-\frac{1}{53}\right)}\right).\frac{123.1010101}{235.1010101}\)

B=\(\frac{47}{41}.\left(\frac{12}{3}:\frac{4}{5}\right).\frac{123}{235}=\frac{47}{41}.\left(\frac{12}{3}.\frac{5}{4}\right).\frac{123}{235}\)

B=\(\frac{47}{41}.\frac{15}{3}.\frac{123}{235}=\frac{47.5.3.41.3}{41.3.5.47}=3\)

Vậy B=3

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Tran Le Khanh Linh
23 tháng 2 2020 lúc 17:20

\(B=1\frac{6}{41}\left(\frac{12+\frac{12}{19}+\frac{12}{37}-\frac{12}{53}}{3+\frac{3}{19}+\frac{3}{27}-\frac{3}{53}}:\frac{4+\frac{4}{19}+\frac{4}{37}-\frac{4}{53}}{5+\frac{5}{19}+\frac{5}{37}-\frac{5}{53}}\right)\cdot\frac{124242423}{237373735}\)

\(B=\frac{47}{6}\cdot\left(\frac{12\left(1+\frac{1}{19}+\frac{1}{37}-\frac{1}{53}\right)}{3\left(1+\frac{1}{19}+\frac{1}{27}-\frac{1}{53}\right)}:\frac{4\left(1+\frac{1}{19}+\frac{1}{37}-\frac{1}{53}\right)}{5\left(1+\frac{1}{19}+\frac{1}{37}-\frac{1}{53}\right)}\right)\cdot\frac{123\cdot1010101}{235\cdot1010101}\)

\(B=\frac{47}{6}\cdot\left(4:\frac{4}{5}\right)\cdot\frac{123}{235}\)

\(B=\frac{47}{6}\cdot5\cdot\frac{123}{235}\)

\(B=3\)

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