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a) \(1-\dfrac{1}{2}=\dfrac{1}{2}\)
\(\dfrac{1}{2}-\dfrac{1}{3}=\dfrac{3-2}{6}=\dfrac{1}{6}\)
\(\dfrac{1}{3}-\dfrac{1}{4}=\dfrac{4-3}{12}=\dfrac{1}{12}\)
\(\dfrac{1}{4}-\dfrac{1}{5}=\dfrac{5-4}{20}=\dfrac{1}{20}\)
\(\dfrac{1}{5}-\dfrac{1}{6}=\dfrac{6-5}{30}=\dfrac{1}{30}\)
b) \(\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}\)
\(=\left(1-\dfrac{1}{2}\right)+\left(\dfrac{1}{2}-\dfrac{1}{3}\right)+\left(\dfrac{1}{3}-\dfrac{1}{4}\right)+\left(\dfrac{1}{4}-\dfrac{1}{5}\right)+\left(\dfrac{1}{5}+\dfrac{1}{6}\right)\)
\(=1+\left(-\dfrac{1}{2}+\dfrac{1}{2}\right)+\left(-\dfrac{1}{3}+\dfrac{1}{3}\right)+\left(-\dfrac{1}{4}+\dfrac{1}{4}\right)+\left(-\dfrac{1}{5}+\dfrac{1}{5}\right)+-\dfrac{1}{6}\)\(=1+-\dfrac{1}{6}\)
\(=\dfrac{5}{6}\)
A, \(\left(\dfrac{8}{15}+\dfrac{14}{23}\right)-\left(\dfrac{5}{15}-\dfrac{9}{23}\right)\)
\(=\dfrac{8}{15}+\dfrac{14}{23}-\dfrac{5}{15}+\dfrac{9}{23}\)
\(=\left(\dfrac{8}{15}-\dfrac{5}{15}\right)+\left(\dfrac{14}{23}+\dfrac{9}{23}\right)\)
\(=\dfrac{3}{15}+1\)
\(=1\dfrac{1}{5}\)
B, \(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+\dfrac{1}{4\cdot5}+\dfrac{1}{5\cdot6}\)
\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}\)
\(=1-\dfrac{1}{6}\)
\(=\dfrac{5}{6}\)
a) \(=\dfrac{8}{15}+\dfrac{14}{23}-\dfrac{5}{15}+\dfrac{9}{23}\)
\(=\dfrac{8}{15}-\dfrac{5}{15}+\dfrac{14}{23}+\dfrac{9}{23}\)
\(=\dfrac{1}{5}+1\)
\(=\dfrac{6}{5}\)
b)
B=\(\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}+\dfrac{1}{56}\)
B=\(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+\dfrac{1}{5.6}+\dfrac{1}{6.7}+\dfrac{1}{7.8}\)
B=\(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}\)
B= 1-\(\dfrac{1}{8}\)
B= \(\dfrac{7}{8}\)
\(A=\dfrac{5}{9}-\dfrac{5}{8}+\dfrac{2}{3}+\dfrac{4}{9}+\dfrac{-3}{8}+\dfrac{1}{3}\\ =\dfrac{5}{9}+\dfrac{-5}{8}+\dfrac{2}{3}+\dfrac{4}{9}+\dfrac{-3}{8}+\dfrac{1}{3}\\= \left(\dfrac{5}{9}+\dfrac{4}{9}\right)+\left(\dfrac{2}{3}+\dfrac{1}{3}\right)+\left(\dfrac{-5}{8}+\dfrac{-3}{8}\right)\\ =1+1+\left(-1\right)\\ =2+\left(-1\right)\\ =1\)
\(-2\dfrac{1}{4}.\)\(\left(3\dfrac{5}{12}-1\dfrac{2}{9}\right)\)
=\(\dfrac{-9}{4}\).\(\left(\dfrac{41}{12}-\dfrac{11}{9}\right)\)
=\(\dfrac{-9}{4}.\dfrac{41}{12}-\dfrac{-9}{4}.\dfrac{11}{9}\)
=\(\dfrac{-123}{16}-\dfrac{-11}{4}\)
=\(\dfrac{-123}{16}-\dfrac{-44}{16}\)
=\(\dfrac{-79}{16}\)
\(\left(-25\%+0,75+\dfrac{7}{12}\right)\div\left(-2\dfrac{1}{8}\right)\)
=\(\left(\dfrac{-1}{4}+\dfrac{3}{4}+\dfrac{7}{12}\right)\div\left(\dfrac{-17}{8}\right)\)
=\(\left(\dfrac{-3}{12}+\dfrac{9}{12}+\dfrac{7}{12}\right).\dfrac{-8}{17}\)
=\(\dfrac{13}{12}.\dfrac{-8}{17}=\dfrac{-26}{51}\)
a: \(=\dfrac{157}{8}\cdot\dfrac{12}{7}-\dfrac{61}{4}\cdot\dfrac{12}{7}\)
\(=\dfrac{12}{7}\left(\dfrac{157}{8}-\dfrac{122}{8}\right)\)
\(=\dfrac{12}{7}\cdot\dfrac{35}{8}=5\cdot\dfrac{3}{2}=\dfrac{15}{2}\)
b: \(=\dfrac{2}{15}-\dfrac{2}{15}\cdot5+\dfrac{3}{15}\)
\(=\dfrac{1}{3}-\dfrac{2}{3}=-\dfrac{1}{3}\)
c: \(=\left(\dfrac{10}{3}+\dfrac{5}{2}\right):\left(\dfrac{19}{6}-\dfrac{21}{5}\right)-\dfrac{11}{31}\)
\(=\dfrac{35}{6}:\dfrac{-31}{30}-\dfrac{11}{31}\)
\(=\dfrac{35}{6}\cdot\dfrac{30}{-31}-\dfrac{11}{31}\)
\(=\dfrac{-35\cdot5-11}{31}=\dfrac{-186}{31}=-6\)
b, \(K =\) \(\dfrac{75}{100}+\dfrac{18}{21}+\dfrac{19}{32}+\dfrac{1}{4}+\dfrac{3}{21}+\dfrac{13}{32}\)
\(K = \) \(\dfrac{3}{4}+\dfrac{18}{21}+\dfrac{19}{32}+\dfrac{1}{4}+\dfrac{3}{21}+\dfrac{13}{32}\)
\(K = \) \(\left(\dfrac{3}{4}+\dfrac{1}{4}\right)+\left(\dfrac{18}{21}+\dfrac{3}{21}\right)+\left(\dfrac{19}{32}+\dfrac{13}{32}\right)\)
\(K = \) \(1 + 1 + 1\)
\(K = \) \(3\)
5, \(\dfrac{1}{7}.\dfrac{2}{5}+\dfrac{1}{7}.\dfrac{1}{5}+\dfrac{1}{7}.\dfrac{4}{5}\)
= \(\dfrac{1}{7}.\left(\dfrac{2}{5}+\dfrac{1}{5}+\dfrac{4}{5}\right)\)
= \(\dfrac{1}{7}\).1
=\(\dfrac{1}{7}\)
5) \(\dfrac{1}{7}.\dfrac{2}{5}+\dfrac{1}{7}.\dfrac{1}{5}+\dfrac{1}{7}.\dfrac{4}{5}\) = \(\dfrac{1}{7}.\left(\dfrac{2}{5}+\dfrac{1}{5}+\dfrac{4}{5}\right)\)
= \(\dfrac{1}{7}.\dfrac{7}{5}=\dfrac{1}{5}\)
a, \(\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{1}{6}+\dfrac{5}{12}+\dfrac{19}{30}\)
\(=\left(\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{5}{12}\right)+\left(\dfrac{1}{5}+\dfrac{1}{6}+\dfrac{19}{30}\right)\)
\(=1+1=2\)
Chúc bạn học tốt!!!
\(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+..+\dfrac{1}{1998.1999}+\dfrac{1}{1999.2000}\)
\(=\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{1998}-\dfrac{1}{1999}+\dfrac{1}{1999}-\dfrac{1}{2000}\)
\(=1-\dfrac{1}{2000}=\dfrac{1999}{2000}.\)