= -103/273

= 0

= -5/7

5.\(-\dfrac{3}{7}+\dfrac{5}{13}+\dfrac{-4}{12}=-\dfrac{103}{273}\)

b.\(-\dfrac{5}{21}+\dfrac{-2}{21}+\dfrac{8}{24}=\dfrac{-5-2}{21}+\dfrac{8}{24}=-\dfrac{7}{21}+\dfrac{8}{24}=-\dfrac{1}{3}+\dfrac{8}{24}=0\)

c.\(\dfrac{5}{13}+\dfrac{-5}{7}+\dfrac{-20}{41}+\dfrac{8}{13}+\dfrac{-21}{41}=\left(\dfrac{5}{13}+\dfrac{8}{13}\right)+\left(\dfrac{-20}{41}+\dfrac{-21}{41}\right)+-\dfrac{5}{7}=1-1-\dfrac{5}{7}=-\dfrac{5}{7}\)

-362/15

Bn có thể giải chi tiết cho mik đc ko ah

\(1\dfrac{13}{15}\times0,75-\left(\dfrac{8}{15}+25\%\right)\)

\(=\dfrac{28}{15}\times\dfrac{3}{4}-\dfrac{8}{15}-25\%\)

\(=\dfrac{7}{5}-\dfrac{8}{15}-\dfrac{1}{4}\)

\(=\dfrac{13}{15}-\dfrac{1}{4}\)

\(=\dfrac{37}{60}\)

\(1\dfrac{13}{15}\cdot0,75-\left[\dfrac{11}{20}+25\%\right]\div\dfrac{7}{5}\\ =\dfrac{28}{15}\cdot0,75-\dfrac{4}{5}\cdot\dfrac{5}{7}\\ =\dfrac{7}{5}-\dfrac{4}{7}=\dfrac{29}{35}\)

 = 4/9 : -8 - (-1/4 + 9/20) : 14/5
= 4/9 : -8 - 1/5 : 14/5
= -1/18 - 1/5 : 14/5
= -1/18 - 1/14
= -8/63

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

\(C=1\)

tick di to lam cho

a, \(\left(\dfrac{3}{8}+\dfrac{-3}{4}+\dfrac{7}{12}\right)\dfrac{5}{6}+\dfrac{1}{2}\)

\(=\left(\dfrac{-3}{8}+\dfrac{7}{12}\right)\dfrac{5}{6}+\dfrac{1}{2}\)

\(=\dfrac{5}{24}.\dfrac{5}{6}+\dfrac{1}{2}\)

\(=\dfrac{25}{144}+\dfrac{1}{2}\)

\(=\dfrac{97}{144}\)

b, \(1\dfrac{13}{15}.0,75-\left(\dfrac{11}{20}+25\%\right):\dfrac{7}{5}\)

\(=\dfrac{28}{15}.0,75-\dfrac{4}{5}:\dfrac{7}{5}\)

\(=\dfrac{7}{5}-\dfrac{4}{7}\)

\(=\dfrac{29}{35}\)

\(A=\dfrac{9-18+14}{24}\cdot\dfrac{6}{5}+\dfrac{1}{2}=\dfrac{5}{24}\cdot\dfrac{6}{5}+\dfrac{1}{2}=\dfrac{1}{4}+\dfrac{1}{2}=\dfrac{3}{4}\)

\(B=\dfrac{28}{15}\cdot\dfrac{3}{4}-\left(\dfrac{11}{20}+\dfrac{1}{4}\right)\cdot\dfrac{5}{7}\)

\(=\dfrac{7}{5}-\dfrac{11+5}{20}\cdot\dfrac{5}{7}\)

\(=\dfrac{7}{5}-\dfrac{4}{7}=\dfrac{49-20}{35}=\dfrac{29}{35}\)

\(\dfrac{1}{2}+\dfrac{3}{4}-\left(\dfrac{3}{4}-\dfrac{4}{5}\right)\\ =\dfrac{1}{2}+\dfrac{3}{4}-\dfrac{3}{4}+\dfrac{4}{5}\\ =\dfrac{1}{2}+\left(\dfrac{3}{4}-\dfrac{3}{4}\right)+\dfrac{4}{5}\\ =\dfrac{1}{2}+0+\dfrac{4}{5}\\ =\dfrac{1}{2}+\dfrac{4}{5}\\ =\dfrac{5}{10}+\dfrac{8}{10}\\ =\dfrac{13}{10}\)

\(\dfrac{-3}{7}+\dfrac{3}{4}:\dfrac{3}{14}\\ =\dfrac{-3}{7}+\dfrac{3}{4}\cdot\dfrac{14}{3}\\ =\dfrac{-3}{7}+\dfrac{7}{2}\\ =\dfrac{-6}{14}+\dfrac{49}{14}\\ =\dfrac{43}{14}\)

\(1\dfrac{13}{15}.0,75-\left(\dfrac{11}{20}+25\%\right):\dfrac{7}{3}\\ =\dfrac{28}{15}.\dfrac{3}{4}-\left(\dfrac{11}{20}+\dfrac{1}{4}\right):\dfrac{7}{3}\\ =\dfrac{7}{5}-\left(\dfrac{11}{20}+\dfrac{5}{20}\right):\dfrac{7}{3}\\ =\dfrac{7}{5}-\dfrac{4}{5}:\dfrac{7}{3}\\ =\dfrac{7}{5}-\dfrac{4}{5}.\dfrac{3}{7}\\ =\dfrac{7}{5}-\dfrac{12}{35}\\ =\dfrac{49}{35}-\dfrac{12}{35}\\ =\dfrac{37}{35}\)