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a) \(\frac{-1}{21}\)\(+\)\(\frac{-1}{28}\)\(=\)\(\frac{-1}{12}\)
b) \(\frac{-8}{18}\)\(-\)\(\frac{15}{27}\)\(=\)\(-1\)
c) \(\frac{-5}{12}\)\(+\)\(0.75\)\(=\)\(\frac{-5}{12}\)\(+\)\(\frac{3}{4}\)\(=\)\(\frac{1}{3}\)
d) \(3.5\)\(-\)\(\left(\frac{-2}{7}\right)\)\(=\)\(\frac{7}{2}\)\(+\)\(\frac{2}{7}\)\(=\)\(\frac{53}{14}\)
\(A=\frac{-5}{12}+\frac{4}{37}+\frac{17}{12}-\frac{41}{37}=(\frac{-5}{12}+\frac{17}{12})+(\frac{4}{37}-\frac{41}{37})=\frac{12}{12}+\frac{-37}{37}=1+(-1)=0\)
\(B=\frac{1}{2}-\frac{43}{101}+\frac{-1}{3}-\frac{1}{6}=\frac{-43}{101}+(\frac{1}{2}+\frac{-1}{3}-\frac{1}{6})=\frac{-43}{101}+(\frac{3}{6}+\frac{-2}{6}-\frac{1}{6})=\frac{-43}{101}+0=\frac{-43}{101}\)
\(A=\frac{-5}{12}+\frac{4}{37}+\frac{17}{12}-\frac{41}{37}.\)
\(A=\left(\frac{-5}{12}+\frac{17}{12}\right)-\left(\frac{41}{37}-\frac{4}{37}\right)\)
\(A=1-1=0\)
\(B=\frac{1}{2}-\frac{43}{101}+\left(\frac{-1}{3}\right)-\frac{1}{6}\)
\(B=\left(\frac{1}{2}+\left(\frac{-1}{3}\right)-\frac{1}{6}\right)-\frac{43}{101}\)
\(A=0-\frac{43}{101}=\frac{-43}{101}\)
\(C=\frac{-5}{6}\cdot\frac{12}{-7}\cdot-\frac{21}{15}\)
\(C=\frac{-5}{2.3}\cdot\frac{3.2.2}{-7}\cdot\frac{3.\left(-7\right)}{3.5}\)
\(C=\frac{-2}{1}=-2\)
\(\left(\frac{-1}{4}+\frac{7}{33}-\frac{5}{3}\right)-\left(\frac{-5}{4}+\frac{6}{11}-\frac{48}{49}\right)=\left(\frac{-1}{4}-\frac{16}{11}\right)-\left(-\frac{31}{44}-\frac{48}{49}\right)=-\frac{1}{4}-\frac{16}{11}+\frac{31}{44}+\frac{48}{49}=-\frac{1}{49}\)
a, \(A=\frac{12}{3.7}+\frac{12}{7.11}+...+\frac{12}{195.199}\)
\(=3.\left(\frac{4}{3.7}+\frac{4}{7.11}+...+\frac{4}{195.199}\right)\)
\(=3.\left(\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+...+\frac{1}{195}-\frac{1}{199}\right)\)
\(=3.\left(\frac{1}{3}-\frac{1}{199}\right)\)
\(=3.\left(\frac{199}{597}-\frac{3}{597}\right)\)
\(=3.\frac{196}{597}\)
\(=\frac{196}{199}\)
a) \(\frac{1}{12}+\frac{3}{15}+\frac{11}{12}+\frac{1}{71}-\frac{12}{10}=\left(\frac{1}{12}+\frac{11}{12}\right)+\left(\frac{1}{5}-\frac{1}{5}\right)+\frac{1}{71}\)
\(=\frac{12}{12}+0+\frac{1}{71}=1+\frac{1}{71}=1\frac{1}{71}=\frac{72}{71}\)
b) \(\frac{2}{3}-4\left(\frac{1}{2}+\frac{3}{4}\right)=\frac{2}{3}-4.\frac{5}{4}=\frac{2}{3}-5=\frac{2}{3}-\frac{15}{3}=-\frac{13}{3}\)
c) \(\frac{-4}{13}.\frac{3}{17}+\frac{-12}{13}.\frac{4}{7}+\frac{4}{13}=\frac{4}{13}.\frac{-3}{17}+\frac{4}{13}.\frac{-12}{17}+\frac{4}{13}.1\)
\(=\frac{4}{13}\left(\frac{-3}{17}+\frac{-12}{17}+1\right)=\frac{4}{13}\left(\frac{-15}{17}+\frac{17}{17}\right)=\frac{4}{13}.\frac{2}{17}=\frac{8}{221}\)
d) \(\frac{10^3+2.5+5^3}{55}=\frac{1000+10+125}{55}=\frac{1135}{55}=\frac{227}{11}\)
\(A=2\left(\frac{1}{30}+\frac{1}{42}+.....+\frac{1}{380}\right)=2\left(\frac{6-5}{5.6}+\frac{7-6}{6.7}+.....+\frac{20-19}{20.19}\right)=2\left(\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+...+\frac{1}{19}-\frac{1}{20}\right)=2\left(\frac{1}{5}-\frac{1}{20}\right)=\frac{3}{10}\)
\(B=\frac{12}{84}+\frac{12}{210}+.....+\frac{12}{2100}=\frac{4}{28}+\frac{4}{70}+.....+\frac{4}{700}=\frac{4}{3}\left(\frac{3}{4.7}+\frac{3}{7.10}+....+\frac{1}{25.28}\right)=\frac{4}{3}\left(\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-.....-\frac{1}{28}\right)=\frac{4}{2}.\frac{6}{28}=\frac{3}{7}\)