\(=\dfrac{\dfrac{1}{3}-\dfrac{1}{7}-\dfrac{1}{13}}{2\left(\dfrac{1}{3}-\dfrac{1}{7}-\dfrac{1}{13}\right)}.\dfrac{\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{5}}{\dfrac{7}{6}-\dfrac{7}{8}+\dfrac{7}{10}}+\dfrac{6}{7}\)

\(=\dfrac{1}{2}.\dfrac{\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{5}}{\dfrac{7}{2}\left(\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{5}\right)}+\dfrac{6}{7}\)

\(=\dfrac{1}{2}.\dfrac{2}{7}+\dfrac{6}{7}=\dfrac{1}{7}+\dfrac{6}{7}=1\)

1)

\(\dfrac{0,375-0,3+\dfrac{3}{11}+\dfrac{3}{12}}{-0,625+0,5-\dfrac{5}{11}-\dfrac{5}{12}}+\dfrac{1,5+1-0,75}{2,5+\dfrac{5}{3}-1,25}\)
\(=\dfrac{\dfrac{3}{8}-\dfrac{3}{10}+\dfrac{3}{11}+\dfrac{3}{12}}{-\dfrac{5}{8}+\dfrac{5}{10}-\dfrac{5}{11}-\dfrac{6}{12}}+\dfrac{\dfrac{3}{2}+\dfrac{3}{3}-\dfrac{3}{4}}{\dfrac{5}{2}+\dfrac{5}{3}-\dfrac{5}{4}}\)
\(=\dfrac{3\left(\dfrac{1}{8}-\dfrac{1}{10}+\dfrac{1}{11}+\dfrac{1}{12}\right)}{-5\left(\dfrac{1}{8}-\dfrac{1}{10}+\dfrac{1}{11}+\dfrac{1}{12}\right)}+\dfrac{3\left(\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}\right)}{5\left(\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}\right)}\)
\(=\dfrac{3}{-5}+\dfrac{3}{5}\)
\(=-\dfrac{3}{5}+\dfrac{3}{5}\)
\(=0\)

\(M=\left(\dfrac{\dfrac{2}{5}-\dfrac{2}{9}+\dfrac{2}{11}}{\dfrac{7}{5}-\dfrac{7}{9}+\dfrac{7}{11}}-\dfrac{\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{5}}{\dfrac{7}{6}-\dfrac{7}{8}+\dfrac{7}{10}}\right)\times\dfrac{2022}{2021}\)

\(M=\left(\dfrac{\dfrac{178}{495}}{\dfrac{623}{495}}-\dfrac{\dfrac{17}{60}}{\dfrac{119}{120}}\right)\times\dfrac{2022}{2021}\)

\(M=\left(\dfrac{2}{7}-\dfrac{2}{7}\right)\times\dfrac{2022}{2021}\)

\(M=0\times\dfrac{2022}{2021}\)

M=0

\(M=\left(\dfrac{0,4-\dfrac{2}{9}+\dfrac{2}{11}}{1,4-\dfrac{7}{9}+\dfrac{7}{11}}-\dfrac{\dfrac{1}{3}-0,25+\dfrac{1}{5}}{1\dfrac{1}{6}-0,875+0,7}\right):\dfrac{2012}{2013}\)

\(M=\left(\dfrac{\dfrac{2}{5}-\dfrac{2}{9}+\dfrac{2}{11}}{\dfrac{7}{5}-\dfrac{7}{9}+\dfrac{7}{11}}-\dfrac{\dfrac{2}{6}-\dfrac{2}{8}+\dfrac{2}{10}}{\dfrac{7}{6}-\dfrac{7}{8}+\dfrac{7}{10}}\right):\dfrac{2012}{2013}\)

\(M=\left(\dfrac{2\left(\dfrac{1}{5}-\dfrac{1}{9}+\dfrac{1}{11}\right)}{7\left(\dfrac{1}{5}-\dfrac{1}{9}+\dfrac{1}{11}\right)}-\dfrac{2\left(\dfrac{1}{6}-\dfrac{1}{8}+\dfrac{1}{10}\right)}{7\left(\dfrac{1}{6}-\dfrac{1}{8}+\dfrac{1}{10}\right)}\right):\dfrac{2012}{2013}\)

\(M=\left(\dfrac{2}{7}-\dfrac{2}{7}\right).\dfrac{2013}{2012}\)

\(M=0.\dfrac{2013}{2012}\)

\(M=0\)

a: \(=\dfrac{9}{13}\cdot\dfrac{4}{5}=\dfrac{36}{65}\)

b: \(=\dfrac{-7}{10}:\dfrac{3}{2}=\dfrac{-7}{10}\cdot\dfrac{2}{3}=\dfrac{-14}{30}=-\dfrac{7}{15}\)

c: \(=\dfrac{7}{6}\left(3+\dfrac{1}{4}-\dfrac{1}{4}\right)=\dfrac{7}{6}\cdot3=\dfrac{7}{2}\)

Bài 1: 

a: \(A=\dfrac{1\left(\dfrac{1}{13}-\dfrac{1}{17}-\dfrac{1}{23}\right)}{2\left(\dfrac{1}{13}-\dfrac{1}{17}-\dfrac{1}{23}\right)}\cdot\dfrac{\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{5}}{\dfrac{7}{6}-\dfrac{7}{8}+\dfrac{7}{10}}+\dfrac{6}{7}\)

\(=\dfrac{1}{2}\cdot\dfrac{2}{7}+\dfrac{6}{7}=\dfrac{1}{7}+\dfrac{6}{7}=1\)

b: \(B=2000:\left[\dfrac{\dfrac{2}{5}-\dfrac{2}{9}+\dfrac{2}{11}}{\dfrac{7}{5}-\dfrac{7}{9}+\dfrac{7}{11}}\cdot\dfrac{-\dfrac{7}{6}+\dfrac{7}{8}-\dfrac{7}{10}}{\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{5}}\right]\)

\(=2000:\left[\dfrac{2}{7}\cdot\dfrac{-7}{2}\right]=-2000\)

c: \(C=10101\cdot\left(\dfrac{5}{111111}+\dfrac{1}{111111}-\dfrac{4}{111111}\right)\)

\(=10101\cdot\dfrac{2}{111111}=\dfrac{2}{11}\)

\(Q=2002:\left[\dfrac{\dfrac{2}{5}-\dfrac{2}{9}+\dfrac{2}{11}}{\dfrac{7}{5}-\dfrac{7}{9}+\dfrac{7}{11}}\cdot\dfrac{\dfrac{-7}{6}+\dfrac{7}{8}-\dfrac{7}{10}}{\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{5}}\right]\)

\(=2002:\left(\dfrac{1}{7}\cdot\dfrac{\dfrac{-7}{2}\left(\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{5}\right)}{\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{5}}\right)\)

\(=2002:\left(\dfrac{1}{7}\cdot\dfrac{-7}{2}\right)=2002:\dfrac{-1}{2}=-4004\)

\(M=\left(\dfrac{0,4-\dfrac{2}{9}+\dfrac{2}{11}}{1,4-\dfrac{7}{9}+\dfrac{7}{11}}-\dfrac{\dfrac{1}{3}-0,25+\dfrac{1}{5}}{1\dfrac{1}{6}-0,875+0,7}\right):\dfrac{2014}{2015}\)

\(=\left[\dfrac{2\left(0,2-\dfrac{1}{9}+\dfrac{1}{11}\right)}{7\left(0,2-\dfrac{1}{9}+\dfrac{1}{11}\right)}-\dfrac{2\left(\dfrac{1}{6}-0,125+\dfrac{1}{10}\right)}{7\left(\dfrac{1}{7}-0,125+\dfrac{1}{10}\right)}\right]:\dfrac{2014}{2015}\)

\(=\left(\dfrac{2}{7}-\dfrac{2}{7}\right):\dfrac{2014}{2015}=0:\dfrac{2014}{2015}=0\)

Vậy M = 0

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