=( 19/13+7/13) + ( 17/9+1/9) + ( 7/3 + 2/3)

= 26/13 + 18/9 + 9/3

= 2 + 2+3

= 7

a) \(-\frac{1}{4}.13\frac{9}{11}-0,25.6\frac{2}{11}\)

\(=-\frac{1}{4}.13\frac{9}{11}-\frac{1}{4}.6\frac{2}{11}\)

\(=-\frac{1}{4}\left(13\frac{9}{11}+6\frac{2}{11}\right)\)

\(=-\frac{1}{4}.20\)

\(=-5\)

b) \(B=\frac{-5}{6}.\frac{4}{19}+\frac{-7}{12}.\frac{4}{19}-\frac{40}{57}\)

\(=\frac{4}{19}\left(\frac{-5}{6}+\frac{-7}{12}\right)-\frac{40}{57}\)

\(=\frac{4}{19}.\frac{-17}{12}-\frac{40}{57}\)

\(=\frac{-17}{57}-\frac{40}{57}\)

\(=-1\)

c)  \(\frac{3}{7}.\frac{9}{26}-\frac{1}{14}.\frac{1}{13}-\frac{1}{7}\)

\(=\frac{3}{7}.\frac{9}{26}-\frac{1}{2}.\frac{1}{7}.\frac{1}{13}-\frac{1}{7}\)

\(=\frac{1}{7}\left(3.\frac{9}{26}-\frac{1}{2}.\frac{1}{13}-1\right)\)

\(=\frac{1}{7}.0\)

\(=0\)

d) \(\frac{4}{9}:\left(-\frac{1}{7}\right)+6\frac{5}{9}:\left(-\frac{1}{7}\right)\)

\(=\left(\frac{4}{9}+6\frac{5}{9}\right):\left(-\frac{1}{7}\right)\)

\(=7:\left(-\frac{1}{7}\right)\)

\(=-49\)

\(A=-\frac{1}{4}x3\frac{9}{11}-0,25x6\frac{2}{11}\)

\(A=-0,25x3\frac{9}{11}-0,25x6\frac{2}{11}\)

\(A=-0,25x\left(\frac{9}{11}+\frac{2}{11}\right)\)

\(A=-0,25x1\)

\(A=-0,25\)

\(B=-\frac{5}{6}x\frac{4}{19}+-\frac{7}{12}x\frac{4}{19}-\frac{40}{57}\)

\(B=-\frac{5}{6}x\frac{4}{19}+-\frac{7}{12}x\frac{4}{19}-\frac{4}{19}x190\)

\(B=\frac{4}{19}x\left(-\frac{5}{6}-\frac{7}{12}-190\right)\)

\(B=\frac{4}{19}x\left(-\frac{2297}{12}\right)\)

\(B=\frac{-2297}{57}\)

\(C=\frac{3}{7}x\frac{9}{26}-\frac{1}{14}x\frac{1}{13}-\frac{1}{7}\)

\(C=\frac{1}{7}x3x\frac{9}{26}-\frac{1}{7}x\frac{1}{2}x\frac{1}{13}-\frac{1}{7}\)

\(C=\frac{1}{7}x\left(3x\frac{9}{26}-\frac{1}{2}x\frac{1}{13}-1\right)\)

\(C=\frac{1}{7}x0\)

\(C=\frac{1}{7}\)

\(D=\frac{4}{9}:\left(-\frac{1}{7}\right)+6\frac{5}{9}:\left(-\frac{1}{7}\right)\)

\(D=\left(\frac{4}{9}+6\frac{5}{9}\right):\left(-\frac{1}{7}\right)\)

\(D=7:\left(-\frac{1}{7}\right)\)

\(D=-49\)

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

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

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

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

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

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

Chúc bn học tốt

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

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

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

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

\(B=3\)

\(\begin{array}{l}\left( {\frac{{20}}{7}.\frac{{ - 4}}{{ - 5}}} \right) + \left( {\frac{{20}}{7}.\frac{3}{{ - 5}}} \right) = \frac{{20}}{7}.\left( {\frac{{ - 4}}{{ - 5}} + \frac{3}{{ - 5}}} \right)\\ = \frac{{20}}{7}.\left( {\frac{{ - 1}}{{ - 5}}} \right) = \frac{{20}}{7}.\frac{1}{5} = \frac{{20}}{{35}} = \frac{4}{7}\end{array}\)

GIÚP ĐỠ MỚI 

nguyễnCôngBắcKỳ_6a1, bài này còn cần nữa k, mik lm cho

Bài làm

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

\(A=\frac{47}{41}.\left(\frac{12+12.\frac{1}{19}+12.\frac{1}{37}-12.\frac{1}{53}}{3+3.\frac{1}{19}+3.\frac{1}{37}-3.\frac{1}{53}}:\frac{4+4.\frac{1}{19}+4.\frac{1}{37}-4.\frac{1}{53}}{5+4.\frac{1}{19}+4.\frac{1}{37}-4.\frac{1}{53}}\right).\frac{124242423}{237373735}\)

\(A=\frac{47}{41}.\left(\frac{12.\left(1+\frac{1}{19}+\frac{1}{37}-\frac{1}{53}\right)}{3.\left(1+\frac{1}{19}+\frac{1}{37}-\frac{1}{53}\right)}:\frac{4.\left(\frac{1}{19}+\frac{1}{37}-\frac{1}{53}\right)+1}{5.4.\left(\frac{1}{19}+\frac{1}{37}-\frac{1}{53}\right)}\right).\frac{124242423}{237373735}\)

\(A=\frac{47}{41}.\left(\frac{12.1}{3.1}:\frac{4.+1}{5.4.}\right).\frac{124242423}{237373735}\)

\(A=\frac{47}{41}.\left(\frac{12}{3}:\frac{5}{20.}\right).\frac{124242423}{237373735}\)

\(A=\frac{47}{41}.\left(\frac{12}{3}x\frac{20}{5}\right).\frac{124242423}{237373735}\)

\(A=\frac{47}{41}.\frac{80}{5}.\frac{124242423}{237373735}\)

\(A=\frac{47}{41}.16.\frac{124242423}{237373735}\)

~ Đến đây tự tính tiếp, bn rút gọn tử phân số \(\frac{124242423}{237373735}\)này thử xem, mik k chắc là rút gọn đc, xg bn tính như bình thường nhé ~
# Chúc bạn học tốt #

Đặt mỗi số số hạng của M là A;B

Ta quy đồng tử và mẫu của A bằng cách nhân cả tử và mẫu với 3.7.13;rồi dùng tính chất phân phối bỏ ngoặc ra.Tính toán bình thường.Khi đó A=1/7 

Tương tự với B ta quy đồng lên với 11.17.19;bỏ ngoặc đi và tính bình thường ;B=2/7

=>M=1/7+2/7-136/7=3/7-136/7=-19