a, \(4\dfrac{5}{37}\)-\(3\dfrac{4}{5}\)+ \(8\dfrac{15}{29}\)- \(3\dfrac{5}{37}\)+ \(6\dfrac{14}{29}\)

=(\(4\dfrac{5}{37}\)-\(3\dfrac{5}{37}\))+(\(8\dfrac{15}{29}\)+\(6\dfrac{14}{29}\))-\(3\dfrac{4}{5}\)

=(4-3)+(\(\dfrac{5}{37}\)-\(\dfrac{5}{37}\))+(8+6)+(\(\dfrac{15}{29}\)+\(\dfrac{14}{29}\))-3\(\dfrac{4}{5}\)

=1+ 15-\(3\dfrac{4}{5}\)=13-\(\dfrac{4}{5}\)=\(\dfrac{61}{5}\)

b, 60\(\dfrac{7}{13}\)+ 50\(\dfrac{8}{13}\)-11\(\dfrac{2}{13}\)

=(60+50-11)+(\(\dfrac{7}{13}\)+ \(\dfrac{8}{13}\)-\(\dfrac{2}{13}\))

=99+1=100

c, đáp án bằng \(\dfrac{-2}{3}\). bạn tự tính nha

\(=2-\left(\dfrac{5}{3}-\dfrac{7}{6}+\dfrac{9}{10}-...-\dfrac{19}{45}\right)\)

\(=2-2\left(\dfrac{5}{6}-\dfrac{7}{12}+\dfrac{9}{20}-...-\dfrac{19}{90}\right)\)

\(=2-2\left(\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}+\dfrac{1}{5}-...-\dfrac{1}{9}-\dfrac{1}{10}\right)\)

\(=2-2\cdot\dfrac{4}{10}=2-\dfrac{8}{10}=2-\dfrac{4}{5}=\dfrac{6}{5}\)

\(=28\left(\dfrac{3}{14}-\dfrac{13}{21}-\dfrac{29}{42}\right)-8\\ =28\cdot\dfrac{-23}{21}-8=-\dfrac{92}{3}-8=-\dfrac{116}{3}\)

\(C=\dfrac{3}{14}:\dfrac{1}{28}-\dfrac{13}{21}:\dfrac{1}{28}-\dfrac{29}{42}:\dfrac{1}{28}-8\)

\(=28\left(\dfrac{3}{14}-\dfrac{13}{21}-\dfrac{29}{42}\right)-8\)

\(=28\left(\dfrac{9}{42}-\dfrac{26}{42}-\dfrac{29}{42}\right)-8\)

\(=28\cdot\dfrac{-46}{42}-8\)

\(=\dfrac{-116}{3}\)

c: \(\dfrac{3}{14}:\dfrac{1}{28}-\dfrac{13}{21}:\dfrac{1}{28}\)

\(=28\left(\dfrac{3}{14}-\dfrac{13}{21}\right)\)

\(=28\cdot\left(\dfrac{9}{42}-\dfrac{26}{42}\right)\)

\(=28\cdot\dfrac{-17}{42}=\dfrac{-34}{3}\)

A= \(\dfrac{1}{3}-\dfrac{3}{5}+\dfrac{5}{7}-\dfrac{7}{9}+\dfrac{9}{11}-\dfrac{5}{7}+\dfrac{3}{5}-\dfrac{9}{11}=\dfrac{1}{3}-\dfrac{7}{9}=\dfrac{3}{9}-\dfrac{7}{9}=-\dfrac{4}{9}\)

\(B=\left(\dfrac{1}{5}+\dfrac{2}{15}+\dfrac{2}{3}\right)+\left(-\dfrac{2}{7}+\dfrac{1}{42}-\dfrac{13}{28}-\dfrac{1}{4}\right)\)

\(=\dfrac{3+2+10}{15}+\dfrac{-2\cdot12+2-13\cdot3-21}{84}\)

=1-82/84

=2/84=1/42

\(C=\dfrac{1}{50}-\left(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{49\cdot50}\right)\)

\(=\dfrac{1}{50}-\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{49}-\dfrac{1}{50}\right)\)

\(=\dfrac{1}{50}-1+\dfrac{1}{50}=\dfrac{1}{25}-1=-\dfrac{24}{25}\)

\(D=\dfrac{3\left(\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{7}+\dfrac{1}{13}\right)}{11\left(\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{7}+\dfrac{1}{13}\right)}=\dfrac{3}{11}\)

1) \(\left(-1\dfrac{1}{5}+x\right):\left(-3\dfrac{3}{5}\right)=\dfrac{-7}{4}+\dfrac{1}{4}:\dfrac{1}{8}\)

\(\Leftrightarrow\left(-1\dfrac{1}{5}+x\right):\left(-3\dfrac{3}{5}\right)=\dfrac{-7}{4}+2\)

\(\Leftrightarrow\left(-1\dfrac{1}{5}+x\right):\left(-3\dfrac{3}{5}\right)=\dfrac{1}{4}\)

\(\Leftrightarrow-1\dfrac{1}{5}+x=\dfrac{1}{4}.\left(-3\dfrac{3}{5}\right)\)

\(\Leftrightarrow-1\dfrac{1}{5}+x=\dfrac{1}{4}.\left(-\dfrac{18}{5}\right)\)

\(\Leftrightarrow-1\dfrac{1}{5}+x=-\dfrac{9}{10}\)

\(\Leftrightarrow x=\left(-\dfrac{9}{10}\right)-\left(-1\dfrac{1}{5}\right)\)

\(\Leftrightarrow x=\dfrac{3}{10}\)

đề có 1/13 ko bn

a: =-5/7(2/11+9/11)+12/7

=12/7-5/7

=7/7=1

b: \(=\dfrac{-12}{56}+\dfrac{35}{56}-\dfrac{28}{56}=\dfrac{-5}{56}\)

c: \(=\dfrac{1}{4}-\dfrac{5}{13}+\dfrac{2}{11}-\dfrac{8}{13}+\dfrac{3}{4}\)

=1-1+2/11

=2/11

d: \(=\dfrac{21}{31}+\dfrac{-16}{7}+\dfrac{44}{53}+\dfrac{10}{31}+\dfrac{9}{53}\)

=1+1-16/7

=-2/7

e: \(=\dfrac{\dfrac{4}{36}-\dfrac{30}{36}-\dfrac{144}{36}}{\dfrac{21}{36}-\dfrac{1}{36}-\dfrac{360}{36}}=\dfrac{-160}{-340}=\dfrac{8}{17}\)