A=-6/5x?viết sai rùi bạn : 4/5

à, bạn j ấy ơi, đề mình đc giao là như thế đấy ạ không nhầm đâu nha!

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

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

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

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

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

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

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

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

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

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

\(B=3\)

A = \(-1\frac{1}{5}.\frac{4\left(3+\frac{1}{3}-\frac{3}{7}-\frac{3}{53}\right)}{3+\frac{1}{3}-\frac{3}{37}-\frac{3}{53}}:\frac{4+\frac{4}{17}+\frac{4}{19}+\frac{4}{2003}}{5+\frac{5}{17}+\frac{5}{19}+\frac{5}{2003}}\)
A = \(-1\frac{1}{5}.\)4 : \(\frac{4.\left(1-\frac{1}{17}+\frac{1}{19}+\frac{1}{2003}\right)}{5.\left(1-\frac{1}{17}+\frac{1}{19}+\frac{1}{2003}\right)}\)
A = \(-1\frac{1}{5}.4\): \(\frac{4}{5}\)= \(\frac{-6}{5}\).4. \(\frac{5}{4}\)
A = \(\frac{-24}{5}.\frac{5}{4}\)=\(\frac{\left(-6\right).1}{1.1}\)= -6.

\(A=-1\frac{1}{5}.\frac{4\left(3+\frac{1}{3}-\frac{3}{37}-\frac{3}{53}\right)}{3+\frac{1}{3}-\frac{3}{37}-\frac{3}{53}}:\frac{4+\frac{4}{17}+\frac{4}{19}+\frac{4}{2003}}{5+\frac{5}{17}+\frac{5}{19}+\frac{5}{2003}}\)

\(=-1\frac{1}{5}.\frac{4\left(3+\frac{1}{3}-\frac{3}{37}-\frac{3}{53}\right)}{3+\frac{1}{3}-\frac{3}{37}-\frac{3}{53}}:\frac{4\left(1+\frac{1}{17}+\frac{1}{19}+\frac{1}{2003}\right)}{5\left(1+\frac{1}{17}+\frac{1}{19}+\frac{1}{2003}\right)}\)

\(=-1\frac{1}{5}.\frac{4}{1}:\frac{4}{5}\)

\(=-1\frac{1}{5}.\frac{4}{1}.\frac{5}{4}\)

\(=-1\)

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 #

\(P=-1\frac{1}{5}.\frac{4\left(3\frac{1}{3}-\frac{3}{37}-\frac{3}{53}\right)}{3+\frac{1}{3}-\frac{3}{37}-\frac{3}{53}}:\frac{4+\frac{4}{17}+\frac{4}{19}+\frac{4}{2003}}{5+\frac{5}{17}+\frac{5}{19}+\frac{5}{2003}}\)

=> \(P=-1\frac{1}{5}.\frac{4\left(3+\frac{1}{3}-\frac{3}{37}-\frac{3}{53}\right)}{3+\frac{1}{3}-\frac{3}{37}-\frac{3}{53}}:\frac{4\left(1+\frac{1}{17}+\frac{1}{19}+\frac{1}{2003}\right)}{5\left(1+\frac{1}{17}+\frac{1}{19}+\frac{1}{2003}\right)}\)

=> \(P=-1\frac{1}{5}.4:\frac{4}{5}\)

=> \(P=-\frac{6}{5}.4.\frac{5}{4}=-6\)

c=39/64

d=913/105

3) C thiếu đề

4) \(D=\frac{1}{9}-\left|\frac{-5}{23}\right|-\left(\frac{-5}{23}+\frac{1}{9}+\frac{25}{7}\right)+\frac{50}{4}-\frac{7}{30}\)

\(D=\frac{1}{9}-\frac{5}{23}+\frac{5}{23}-\frac{1}{9}-\frac{25}{7}+\frac{50}{4}-\frac{7}{30}\)

\(D=\frac{1}{9}-\frac{1}{9}-\frac{5}{23}+\frac{5}{23}+\frac{-25}{7}+\frac{50}{4}-\frac{7}{30}\)

\(D=0+0+\frac{125}{14}-\frac{7}{30}\)

\(D=\frac{913}{105}\)

\(C=\frac{5}{18}+\frac{8}{19}-\frac{7}{21}+\left(\frac{-10}{36}+\frac{11}{19}+\frac{1}{3}\right)-\frac{5}{8}\)

\(C=\frac{5}{18}+\frac{8}{19}-\frac{1}{3}+\frac{-5}{18}+\frac{11}{19}+\frac{1}{3}-\frac{5}{8}\)

\(C=\left(\frac{5}{18}+\frac{-5}{18}\right)+\left(\frac{8}{19}+\frac{11}{19}\right)+\left(\frac{-1}{3}+\frac{1}{3}\right)-\frac{5}{8}\)

\(C=0+1+0-\frac{5}{8}\)

\(C=\frac{3}{8}\)