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\(E=\dfrac{\left(\dfrac{53}{4}-\dfrac{59}{27}-\dfrac{65}{6}\right).\dfrac{5751}{25}+\dfrac{187}{4}}{\left(\dfrac{10}{7}+\dfrac{10}{3}\right):\left(\dfrac{37}{3}-\dfrac{100}{7}\right)}\)
\(=\dfrac{\dfrac{25}{108}.\dfrac{5751}{25}+\dfrac{187}{4}}{\dfrac{100}{21}:\left(\dfrac{-44}{21}\right)}\)
\(=\dfrac{53,25+\dfrac{187}{4}}{\dfrac{-25}{11}}\)
\(=\dfrac{100}{\dfrac{-25}{11}}\)
\(=-44\)
\(E=\dfrac{\left(13\dfrac{1}{4}-2\dfrac{5}{27}-10\dfrac{5}{6}\right).230\dfrac{1}{25}+46\dfrac{3}{4}}{\left(1\dfrac{3}{7}+\dfrac{10}{3}\right):\left(12\dfrac{1}{3}-14\dfrac{2}{7}\right)}\)
\(E=\dfrac{\left(\dfrac{53}{4}-\dfrac{59}{27}-\dfrac{65}{6}\right).\dfrac{5751}{25}+\dfrac{187}{4}}{\dfrac{100}{21}:\left(-\dfrac{41}{21}\right)}\)
\(E=\dfrac{\dfrac{25}{108}.\dfrac{5751}{25}+\dfrac{187}{4}}{-\dfrac{100}{41}}\)
\(E=\dfrac{\dfrac{213}{4}+\dfrac{187}{4}}{-\dfrac{100}{41}}\)
\(E=\dfrac{100}{-\dfrac{100}{41}}\)
\(E=-41\)
\(=\dfrac{\left(\dfrac{53}{4}-\dfrac{59}{27}-\dfrac{65}{6}\right)\cdot230.04+46.75}{\left(\dfrac{13}{10}+\dfrac{10}{3}\right):\left(\dfrac{37}{3}-\dfrac{100}{7}\right)}\)
\(=\dfrac{53.25+46.75}{\dfrac{139}{30}:\dfrac{-41}{21}}=100:\left(\dfrac{139}{30}\cdot\dfrac{-21}{41}\right)\)
\(=100\cdot\dfrac{-400}{973}=\dfrac{-40000}{973}\)
\(\dfrac{\left(13\dfrac{1}{4}-2\dfrac{5}{27}-10\dfrac{5}{6}\right).203\dfrac{1}{25}+46\dfrac{3}{4}}{\left(1\dfrac{3}{7}+\dfrac{10}{3}\right):\left(12\dfrac{1}{3}-14\dfrac{2}{7}\right)}\)
= \(\dfrac{\left(\dfrac{53}{4}-\dfrac{59}{27}-\dfrac{65}{6}\right).\dfrac{5076}{25}+\dfrac{187}{4}}{\left(\dfrac{10}{7}+\dfrac{10}{3}\right):\left(\dfrac{37}{3}-\dfrac{100}{7}\right)}\)
= \(\dfrac{\dfrac{25}{108}.\dfrac{5076}{25}+\dfrac{187}{4}}{\dfrac{100}{21}:\dfrac{-41}{21}}\)
=\(\dfrac{\dfrac{5076}{108}+\dfrac{187}{4}}{\dfrac{-100}{4}}\)
= \(\dfrac{47+\dfrac{187}{4}}{-25}\)
= \(\dfrac{375}{4}:\left(-25\right)\)
= \(\dfrac{-15}{4}\)
1: \(A=\dfrac{-25}{27}-\dfrac{31}{42}+\dfrac{7}{27}+\dfrac{3}{42}=\dfrac{-2}{3}-\dfrac{2}{3}=\dfrac{-4}{3}\)
2: \(B=\dfrac{10.3-\left(9.5-4.5\right)\cdot2}{1.2-1.5}=\dfrac{10.3-10}{-0.3}=-1\)
c: \(=\dfrac{3}{49}\left(\dfrac{19}{2}-\dfrac{5}{2}\right)-\left(\dfrac{1}{20}-\dfrac{5}{20}\right)^2\cdot\left(\dfrac{-7}{14}-\dfrac{193}{14}\right)\)
\(=\dfrac{3}{49}\cdot7-\dfrac{1}{25}\cdot\dfrac{-200}{14}\)
\(=\dfrac{3}{7}+\dfrac{8}{14}=1\)