\(1,A=-\dfrac{3}{4}.\left(0,125-1\dfrac{1}{2}\right):\dfrac{33}{16}-25\%\)

\(A=-\dfrac{3}{4}.\left(0,125-\dfrac{3}{2}\right):\dfrac{33}{16}-\dfrac{1}{4}\)

\(A=-\dfrac{3}{4}.\left(-\dfrac{11}{8}\right):\dfrac{33}{16}-\dfrac{1}{4}\)

\(A=\dfrac{33}{32}:\dfrac{33}{16}-\dfrac{1}{4}\)

\(A=\dfrac{33}{32}.\dfrac{16}{33}-\dfrac{1}{4}\)

\(A=\dfrac{1}{2}-\dfrac{1}{4}\)

\(A=\dfrac{2}{4}-\dfrac{1}{4}\)

\(A=\dfrac{1}{4}\)

\(D=6\dfrac{5}{12}:2\dfrac{5}{4}+11\dfrac{1}{4}.\left(\dfrac{1}{3}-\dfrac{1}{5}\right)\)

\(D=\dfrac{77}{12}:\dfrac{13}{4}+\dfrac{45}{4}.\dfrac{2}{15}\)

\(D=\dfrac{77}{39}+\dfrac{3}{2}\)

\(D=\dfrac{271}{78}\)

\(C=\dfrac{5}{16}:0,125-\left(2\dfrac{1}{4}-0,6\right).\dfrac{10}{11}\)

\(C=\dfrac{5}{16}:0,125-\left(\dfrac{9}{4}-0,6\right).\dfrac{10}{11}\)

\(C=\dfrac{5}{16}:0,125-\dfrac{33}{20}.\dfrac{10}{11}\)

\(C=\dfrac{5}{2}-\dfrac{3}{2}\)

\(C=1\)

a) \(\left(1,25\right)^3.8^3=\left(1,25.8\right)^3=1000\)

b) \(\left(\dfrac{-11}{9}\right)^4.\left(\dfrac{27}{22}\right)^4=\left(\dfrac{-11}{9}.\dfrac{27}{22}\right)^4=\left(\dfrac{-11.9.3}{9.2.\left(-11\right)}\right)^4\)

\(=\left(\dfrac{3}{2}\right)^4=\dfrac{81}{16}\)

c) \(\left(\dfrac{3}{7}+\dfrac{1}{2}\right)^2=\left(\dfrac{13}{14}\right)^2=\dfrac{169}{196}\)

d) \(\dfrac{5^4.20^4}{25^5.4^5}=\dfrac{\left(5.20\right)^4}{\left(25.4\right)^5}=\dfrac{100^4}{100^5}=100^{-1}=0,01\)

a) (-5) . (-4) + (-5) . 14 = (-5 ) . [ (-4) + 14 ] = -50

b ) 3 . ( 5 + 8 ) = 13 . ( -3 ) + 13 . 8 = 65

a) \(5\dfrac{4}{23}.27\dfrac{3}{47}+4\dfrac{3}{47}.\left(-5\dfrac{4}{23}\right)\)

\(=5\dfrac{4}{23}.27\dfrac{3}{47}+\left(-4\dfrac{3}{47}\right).5\dfrac{4}{23}\)

\(=5\dfrac{4}{23}.\left[27\dfrac{3}{47}+\left(-4\dfrac{3}{47}\right)\right]\)

\(=5\dfrac{4}{23}.\left(27\dfrac{3}{47}-4\dfrac{3}{27}\right)\)

\(=5\dfrac{4}{23}.23\)

\(=\dfrac{119}{23}.23\)

\(=\dfrac{119}{23}\)

b) \(4.\left(\dfrac{-1}{2}\right)^3+\dfrac{3}{2}\)

\(=4.\dfrac{-1}{6}+\dfrac{3}{2}\)

\(=\dfrac{-4}{6}+\dfrac{3}{2}\)

\(=\dfrac{-2}{3}+\dfrac{3}{2}\)

\(=\dfrac{-4}{6}+\dfrac{9}{6}\)

\(=\dfrac{5}{6}\)

c) \(\left(\dfrac{1999}{2011}-\dfrac{2011}{1999}\right)-\left(\dfrac{-12}{1999}-\dfrac{12}{2011}\right)\)

\(=\dfrac{1999}{2011}-\dfrac{2011}{1999}-\dfrac{-12}{1999}+\dfrac{12}{2011}\)

\(=\left(\dfrac{1999}{2011}+\dfrac{12}{2011}\right)-\left(\dfrac{2011}{1999}+\dfrac{-12}{1999}\right)\)

\(=\dfrac{2011}{2011}-\dfrac{1999}{1999}\)

\(=1-1\)

\(=0\)

d) \(\left(\dfrac{-5}{11}+\dfrac{7}{22}-\dfrac{-4}{33}-\dfrac{5}{44}\right):\left(\dfrac{381}{22}-39\dfrac{7}{22}\right)\)

(đợi đã, mình chưa tìm được hướng làm...)

d) \(\left(\dfrac{-5}{11}+\dfrac{7}{22}-\dfrac{-4}{33}-\dfrac{5}{44}\right):\left(\dfrac{381}{22}-39\dfrac{7}{22}\right)\)

\(=\left(\dfrac{-60}{132}+\dfrac{42}{132}-\dfrac{-16}{132}-\dfrac{15}{132}\right):\left(\dfrac{381}{22}-39\dfrac{7}{22}\right)\)

\(=\dfrac{-17}{132}:\left(\dfrac{381}{22}-\dfrac{865}{22}\right)\)

\(=\dfrac{-17}{132}:\left(-22\right)\)

\(=\dfrac{-17}{132}.\dfrac{1}{-22}\)

\(=\dfrac{-17}{-2904}=\dfrac{17}{2904}\)

hơi sai sai tki pGiang Thủy Tiên

sai đề r bạn ơiiiii

a)=-1000000

b)=81/56

chúc bạn học tốt

\(a,=\dfrac{13}{50}\cdot\dfrac{50}{13}\cdot\left(-\dfrac{31}{2}\right)\cdot\dfrac{169}{2}=-\dfrac{5239}{2}\\ b,=\dfrac{-\dfrac{49}{100}\cdot\left(-125\right)}{-\dfrac{343}{27}\cdot\dfrac{81}{16}\cdot\left(-1\right)}=\dfrac{\dfrac{245}{4}}{\dfrac{1029}{16}}=\dfrac{245}{4}\cdot\dfrac{16}{1029}=\dfrac{20}{21}\)

a) \(\dfrac{13}{50}.\left(-15.5\right):\dfrac{13}{50}.84\dfrac{1}{2}=\dfrac{13}{50}.-75:\dfrac{13}{50}.\dfrac{169}{2}=-\dfrac{75.169}{2}=-\dfrac{12675}{2}\)

b) \(\dfrac{\left(-0,7\right)^2.\left(-5\right)^3}{\left(-2\dfrac{1}{3}\right)^3.\left(1\dfrac{1}{2}\right)^4.\left(-1\right)^5}=\dfrac{0,49.\left(-125\right)}{-\dfrac{343}{27}.\dfrac{81}{16}.\left(-1\right)}=-\dfrac{\dfrac{245}{4}}{\dfrac{1029}{16}}=\dfrac{20}{21}\)