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\(E=\dfrac{-1}{3}-\dfrac{2}{3}+\dfrac{2}{5}-\dfrac{3}{5}+\dfrac{1}{5}=-1\)
\(\dfrac{7}{21}+\dfrac{-9}{36}=\dfrac{1}{3}+\dfrac{-1}{4}=\dfrac{4}{12}+\dfrac{-3}{12}=\dfrac{1}{12}\)
\(\dfrac{-12}{18}+\dfrac{-21}{35}=\dfrac{-2}{3}+\dfrac{-3}{5}=\dfrac{-10}{15}+\dfrac{-9}{15}=\dfrac{-19}{15}\)
\(\dfrac{-18}{14}+\dfrac{15}{-21}=\dfrac{-9}{7}+\dfrac{-5}{7}=\dfrac{-14}{7}=-2\)
\(\dfrac{3}{21}+\dfrac{-6}{42}=\dfrac{1}{7}+\dfrac{-1}{7}=0\)
a) A= \(\dfrac{4}{7}+\dfrac{3}{4}+\dfrac{2}{7}+\dfrac{5}{4}+\dfrac{1}{7}\)
= \(\left(\dfrac{4}{7}+\dfrac{2}{7}+\dfrac{1}{7}\right)+\left(\dfrac{3}{4}+\dfrac{5}{4}\right)\)
= \(\dfrac{4+2+1}{7}+\dfrac{3+5}{4}\)
= \(\dfrac{7}{7}+\dfrac{8}{4}\) = \(1+2\) = \(3\)
b) B= \(\dfrac{-4}{12}+\dfrac{18}{45}+\dfrac{-6}{9}+\dfrac{-21}{35}+\dfrac{6}{30}\)
= \(\dfrac{-1}{3}+\dfrac{2}{5}+\dfrac{-2}{3}+\dfrac{-3}{5}+\dfrac{1}{5}\)
= \(\left(\dfrac{-1}{3}+\dfrac{-2}{3}\right)+\left(\dfrac{2}{5}+\dfrac{-3}{5}+\dfrac{1}{5}\right)\)
= \(\dfrac{\left(-1\right)+\left(-2\right)}{3}+\dfrac{2+\left(-3\right)+1}{5}\)
= \(\dfrac{-3}{3}+\dfrac{0}{5}\) = \(-1+0\) = \(-1\)
1: \(\dfrac{1}{2}+\dfrac{9}{10}+\dfrac{5}{6}-\dfrac{11}{14}-\dfrac{1}{3}+\dfrac{-4}{35}\)
\(=\left(\dfrac{1}{2}+\dfrac{5}{6}-\dfrac{1}{3}\right)+\dfrac{9}{10}-\left(\dfrac{11}{14}+\dfrac{4}{35}\right)\)
\(=\dfrac{3+5-2}{6}+\dfrac{9}{10}-\dfrac{55+8}{70}\)
\(=1+\dfrac{9}{10}-\dfrac{9}{10}\)
=1
a) Ta có: \(\dfrac{-5}{18}+\dfrac{32}{45}-\dfrac{9}{10}\)
\(=\dfrac{-25}{90}+\dfrac{64}{90}-\dfrac{81}{90}\)
\(=\dfrac{-42}{90}=-\dfrac{7}{15}\)
b) Ta có: \(\left(-\dfrac{1}{4}+\dfrac{51}{33}-\dfrac{5}{3}\right)-\left(-\dfrac{15}{12}+\dfrac{6}{11}-\dfrac{42}{29}\right)\)
\(=\dfrac{-1}{4}+\dfrac{17}{11}-\dfrac{5}{3}+\dfrac{5}{4}-\dfrac{6}{11}+\dfrac{42}{29}\)
\(=\dfrac{-5}{3}+\dfrac{42}{29}\)
\(=\dfrac{-145}{87}+\dfrac{126}{87}=\dfrac{-19}{87}\)
c) Ta có: \(1-\dfrac{1}{2}+2-\dfrac{2}{3}+3-\dfrac{3}{4}+4-\dfrac{1}{4}-3-\dfrac{1}{3}-2-\dfrac{1}{2}-1\)
\(=\left(1-1\right)-\left(\dfrac{1}{2}+\dfrac{1}{2}\right)+\left(2-2\right)-\left(\dfrac{2}{3}+\dfrac{1}{3}\right)+\left(3-3\right)-\left(\dfrac{3}{4}+\dfrac{1}{4}\right)+4\)
\(=-1-1-1+4\)
=1
a) Ta có: −518+3245−910−518+3245−910
=−2590+6490−8190=−2590+6490−8190
=−4290=−715=−4290=−715
b) Ta có: (−14+5133−53)−(−1512+611−4229)(−14+5133−53)−(−1512+611−4229)
=−14+1711−53+54−611+4229=−14+1711−53+54−611+4229
=−53+4229=−53+4229
=−14587+12687=−1987=−14587+12687=−1987
c) Ta có: 1−12+2−23+3−34+4−14−3−13−2−12−11−12+2−23+3−34+4−14−3−13−2−12−1
=(1−1)−(12+12)+(2−2)−(23+13)+(3−3)−(34+14)+4=(1−1)−(12+12)+(2−2)−(23+13)+(3−3)−(34+14)+4
=−1−1−1+4=−1−1−1+4
=1
\(\dfrac{2}{3}+\dfrac{5}{6}+\dfrac{9}{10}+\dfrac{14}{15}+\dfrac{20}{21}+\dfrac{27}{28}+\dfrac{35}{36}+\dfrac{44}{45}\\ =\left(1-\dfrac{1}{3}\right)+\left(1-\dfrac{1}{6}\right)+\left(1-\dfrac{1}{10}\right)+\left(1-\dfrac{1}{15}\right)+\left(1-\dfrac{1}{21}\right)+\left(1-\dfrac{1}{28}\right)+\left(1-\dfrac{1}{36}\right)+\left(1-\dfrac{1}{45}\right)\\ =8-\left(\dfrac{1}{3}+\dfrac{1}{6}+\dfrac{1}{10}+\dfrac{1}{15}+\dfrac{1}{21}+\dfrac{1}{28}+\dfrac{1}{36}+\dfrac{1}{45}\right)\\ =8-\left(\dfrac{2}{6}+\dfrac{2}{12}+\dfrac{2}{20}+\dfrac{2}{30}+\dfrac{2}{42}+\dfrac{2}{56}+\dfrac{1}{72}+\dfrac{1}{90}\right)\\ =8-\left(\dfrac{2}{2.3}+\dfrac{2}{3.4}+\dfrac{2}{4.5}+\dfrac{2}{5.6}+\dfrac{2}{6.7}+\dfrac{2}{7.8}+\dfrac{2}{8.9}+\dfrac{2}{9.10}\right)\\ =8-2\left(\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+\dfrac{1}{5.6}+\dfrac{1}{6.7}+\dfrac{1}{7.8}+\dfrac{1}{8.9}+\dfrac{1}{9.10}\right)\\ =8-2\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{10}\right)\\ =8-2\left(\dfrac{1}{2}-\dfrac{1}{10}\right)=8-2.\dfrac{2}{5}=8-\dfrac{4}{5}=\dfrac{36}{5}\)
Ta có: \(\dfrac{-4}{12}+\dfrac{18}{45}+\dfrac{-6}{9}+\dfrac{21}{35}+\dfrac{6}{30}\)
\(=\dfrac{-1}{3}+\dfrac{-2}{3}+\dfrac{2}{5}+\dfrac{3}{5}+\dfrac{6}{30}\)
\(=-1+1+\dfrac{1}{5}\)
\(=\dfrac{1}{5}\)