\(I=\dfrac{5}{4}+\dfrac{-1}{3}-\dfrac{5}{-24}=\dfrac{9}{8}\)

\(J=\dfrac{-19}{-9}+\dfrac{4}{-11}-\dfrac{-2}{3}=\dfrac{239}{99}\)

\(K=\dfrac{-5}{6}-\dfrac{7}{12}+\dfrac{-3}{4}=-\dfrac{13}{6}\)

\(L=\dfrac{-3}{20}+\dfrac{1}{5}-\dfrac{-5}{3}=\dfrac{103}{60}\)

1/ \(\left(\dfrac{2021}{2020}+\dfrac{2020}{2021}\right).\left(\dfrac{1}{2}-\dfrac{1}{3}-\dfrac{1}{6}\right)\)

=\(\left(\dfrac{2021}{2020}+\dfrac{2020}{2021}\right).0\)

=\(0\)

mink chịu bài này nó rất khó

1) \(\dfrac{1}{2}+\dfrac{13}{19}-\dfrac{4}{9}+\dfrac{6}{19}+\dfrac{5}{18}\)

\(=\dfrac{1}{2}+\left(\dfrac{13}{19}+\dfrac{6}{19}\right)-\dfrac{4}{9}+\dfrac{5}{18}\)

\(=\dfrac{3}{2}-\dfrac{4}{9}+\dfrac{5}{18}\)

\(=\dfrac{19}{18}+\dfrac{5}{18}\)

\(=\dfrac{24}{18}\)

\(=\dfrac{4}{3}\)

2) \(\dfrac{-20}{23}+\dfrac{2}{3}-\dfrac{3}{23}+\dfrac{2}{5}+\dfrac{7}{15}\)

\(=\left(-\dfrac{20}{23}-\dfrac{3}{23}\right)+\dfrac{2}{3}+\dfrac{2}{5}+\dfrac{7}{15}\)

\(=-1+\dfrac{2}{3}+\dfrac{2}{5}+\dfrac{7}{15}\)

\(=-\dfrac{1}{3}+\dfrac{2}{5}+\dfrac{7}{15}\)

\(=\dfrac{1}{15}+\dfrac{7}{15}\)

\(=\dfrac{8}{15}\)

3) \(\dfrac{4}{3}+\dfrac{-11}{31}+\dfrac{3}{10}-\dfrac{20}{31}-\dfrac{2}{5}\)

\(=\left(\dfrac{-11}{31}-\dfrac{20}{31}\right)+\dfrac{4}{3}+\dfrac{3}{10}-\dfrac{2}{5}\)

\(=-1+\dfrac{4}{3}+\dfrac{3}{10}-\dfrac{2}{5}\)

\(=\dfrac{1}{3}+\dfrac{3}{10}-\dfrac{2}{5}\)

\(=\dfrac{1}{3}-\dfrac{1}{10}\)

\(=\dfrac{7}{30}\)

4) \(\dfrac{5}{7}.\dfrac{5}{11}+\dfrac{5}{7}.\dfrac{2}{11}-\dfrac{5}{7}.\dfrac{14}{11}\)

\(=\dfrac{5}{7}.\left(\dfrac{5}{11}+\dfrac{2}{11}-\dfrac{14}{11}\right)\)

\(=\dfrac{5}{7}.-\dfrac{7}{11}\)

\(=-\dfrac{35}{77}\)

\(=-\dfrac{5}{11}\)

Ta có:B=1\(\dfrac{6}{41}\)( \(\dfrac{12+\dfrac{12}{19}-\dfrac{12}{37}-\dfrac{12}{53}}{3+\dfrac{1}{3}-\dfrac{3}{37}-\dfrac{3}{53}}:\dfrac{4+\dfrac{4}{17}+\dfrac{4}{19}+\dfrac{4}{2006}}{5+\dfrac{5}{17}+\dfrac{5}{19}+\dfrac{5}{2006}}\) )

B=\(\dfrac{47}{41}\) [\(\dfrac{12\left(1+\dfrac{1}{19}-\dfrac{1}{37}-\dfrac{1}{53}\right)}{3\left(1+\dfrac{1}{3}-\dfrac{1}{37}-\dfrac{1}{53}\right)}:\dfrac{4\left(\dfrac{1}{17}+\dfrac{1}{19}+\dfrac{1}{2006}\right)}{5\left(1+\dfrac{1}{17}+\dfrac{1}{19}+\dfrac{1}{2006}\right)}\) B = \(\dfrac{47}{41}\) [ \(\dfrac{12}{3}:\dfrac{4}{5}\)]

B = \(\dfrac{47}{41}\)[ 4 . \(\dfrac{5}{4}\)]

B = \(\dfrac{47}{41}.5\)

B = \(\dfrac{235}{41}\)

Chúc bn hc tốt!!!

    Mới thế đã hai năm trôi qua,câu trả lời từ mọi người vẫn KO XUẤT HIỆN.

    Ko biết sau này câu trả lời có xuất hiện hay ko...

g: \(=\dfrac{-3}{4}-\dfrac{1}{4}+\dfrac{5}{7}+\dfrac{2}{7}+\dfrac{3}{5}=\dfrac{3}{5}\)

h: \(=\dfrac{7}{19}\left(\dfrac{8}{11}+\dfrac{3}{11}\right)-\dfrac{12}{19}=\dfrac{7}{19}-\dfrac{12}{19}=-\dfrac{5}{19}\)

i: \(=\dfrac{2013}{7}\left(19+\dfrac{5}{8}-26-\dfrac{5}{8}\right)=\dfrac{2013}{7}\cdot\left(-7\right)=-2013\)

a: \(A=\dfrac{19}{9}+\dfrac{4}{11}+\dfrac{2}{3}=\dfrac{209}{99}+\dfrac{44}{99}+\dfrac{66}{99}=\dfrac{319}{99}\)

b: \(B=\dfrac{-50}{60}+\dfrac{-35}{60}+\dfrac{12}{60}=\dfrac{-73}{60}\)

c: \(C=\dfrac{-27}{36}+\dfrac{132}{36}+\dfrac{10}{36}=\dfrac{115}{36}\)

d: \(D=\dfrac{-19}{3}+\dfrac{2}{3}-\dfrac{4}{5}=\dfrac{-17}{3}-\dfrac{4}{5}=\dfrac{-85-12}{15}=-\dfrac{97}{15}\)

\(-1\dfrac{1}{5}.\dfrac{12+\dfrac{4}{3}-\dfrac{12}{37}-\dfrac{12}{35}}{3+\dfrac{1}{3}-\dfrac{3}{37}-\dfrac{3}{35}}:\dfrac{4+\dfrac{4}{17}+\dfrac{4}{19}+\dfrac{4}{2003}}{5+\dfrac{5}{17}+\dfrac{5}{19}+\dfrac{5}{2003}}\)

\(=\dfrac{-6}{5}.\dfrac{4\left(3+\dfrac{1}{3}-\dfrac{3}{37}-\dfrac{3}{35}\right)}{3+\dfrac{1}{3}-\dfrac{3}{37}-\dfrac{3}{35}}:\dfrac{4\left(1+\dfrac{1}{17}+\dfrac{1}{19}+\dfrac{1}{2003}\right)}{5\left(1+\dfrac{1}{17}+\dfrac{1}{19}+\dfrac{1}{2003}\right)}\)

\(=\dfrac{-6}{5}.4:\dfrac{4}{5}\)

\(=\dfrac{-6.4.5}{5.4}=-6\)

= -4 đúng không

\(-1\dfrac{1}{5}.\dfrac{12+\dfrac{4}{3}-\dfrac{12}{37}-\dfrac{12}{35}}{3+\dfrac{1}{3}-\dfrac{3}{37}-\dfrac{3}{53}}:\dfrac{4+\dfrac{4}{17}+\dfrac{4}{19}+\dfrac{4}{2003}}{5+\dfrac{5}{17}+\dfrac{5}{19}+\dfrac{5}{2003}}\)

\(=\dfrac{-6}{5}.\dfrac{4\left(3+\dfrac{1}{3}-\dfrac{3}{37}-\dfrac{3}{53}\right)}{3+\dfrac{1}{3}-\dfrac{3}{37}-\dfrac{3}{53}}:\dfrac{4\left(1+\dfrac{1}{17}+\dfrac{1}{19}+\dfrac{1}{2003}\right)}{5\left(1+\dfrac{1}{17}+\dfrac{1}{19}+\dfrac{1}{2003}\right)}\)

\(=\dfrac{-6}{5}.\dfrac{4}{3}:\dfrac{4}{5}\)

\(=\dfrac{-6.4.5}{5.3.4}=\dfrac{-6}{3}=-2\)

Vậy...