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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)
a/ \(\dfrac{3}{11.12}+\dfrac{3}{12.13}+\dfrac{3}{13.14}+\dfrac{3}{14.15}\)
\(=3\left(\dfrac{1}{11.12}+\dfrac{1}{12.13}+\dfrac{1}{13.14}+\dfrac{1}{14.15}\right)\)
\(=3\left(\dfrac{1}{11}-\dfrac{1}{12}+\dfrac{1}{12}-\dfrac{1}{13}+\dfrac{1}{13}-\dfrac{1}{14}+\dfrac{1}{14}-\dfrac{1}{15}\right)\)
\(=3\left(\dfrac{1}{11}-\dfrac{1}{15}\right)\)
\(=\dfrac{4}{55}\)
b/ \(\dfrac{2}{2.3}+\dfrac{2}{3.4}+\dfrac{2}{4.5}+\dfrac{2}{5.6}\)
\(=2\left(\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+\dfrac{1}{5.6}\right)\)
\(=2\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}\right)\)
\(=2\left(\dfrac{1}{2}-\dfrac{1}{6}\right)\)
\(=\dfrac{2}{3}\)
c/ \(\dfrac{3}{1.4}+\dfrac{3}{4.7}+.....+\dfrac{3}{97.100}\)
\(=1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+....+\dfrac{1}{97}-\dfrac{1}{100}\)
\(=1-\dfrac{1}{100}\)
\(=\dfrac{99}{100}\)
d/ \(\dfrac{3}{2.5}+\dfrac{3}{5.8}+.....+\dfrac{3}{100.103}\)
\(=\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+....+\dfrac{1}{100}-\dfrac{1}{103}\)
\(=\dfrac{1}{2}-\dfrac{1}{103}\)
\(=\dfrac{101}{206}\)
e/ Đặt :
\(A=\dfrac{1}{1.5}+\dfrac{1}{5.10}+....+\dfrac{1}{95.100}\)
\(\Leftrightarrow5A=\dfrac{5}{1.5}+\dfrac{5}{5.10}+....+\dfrac{5}{95.100}\)
\(=1-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{10}+....+\dfrac{1}{95}-\dfrac{1}{100}\)
\(=1-\dfrac{1}{100}\)
\(=\dfrac{99}{100}\)
\(\Leftrightarrow A=\dfrac{99}{100}:5=\dfrac{99}{500}\)
Dấu . là dấu nhân nhé <3
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) Tinh nhanh:
a) \(\dfrac{5}{23}\) . \(\dfrac{17}{26}\) + \(\dfrac{5}{23}\) . \(\dfrac{10}{26}\) - \(\dfrac{5}{23}\)
= \(\dfrac{5}{23}\) . \(\left(\dfrac{17}{26}+\dfrac{10}{26}-1\right)\)
= \(\dfrac{5}{23}\) . \(\left(\dfrac{27}{26}-1\right)\) = \(\dfrac{5}{23}\) . \(\dfrac{1}{26}\)
= \(\dfrac{5}{598}\)
b) \(\dfrac{1}{7}.\dfrac{5}{9}+\dfrac{5}{9}.\dfrac{2}{7}+\dfrac{5}{9}.\dfrac{1}{7}+\dfrac{5}{9}.\dfrac{3}{7}\)
= \(\dfrac{5}{9}.\left(\dfrac{1}{7}+\dfrac{2}{7}+\dfrac{1}{7}+\dfrac{3}{7}\right)\)
= \(\dfrac{5}{9}\) . 1= \(\dfrac{5}{9}\)
A=\(\dfrac{2}{7}+\dfrac{-3}{8}+\dfrac{11}{7}+\dfrac{1}{3}+\dfrac{1}{7}+\dfrac{5}{-3}\)
A=\(\left(\dfrac{2}{7}+\dfrac{11}{7}+\dfrac{1}{7}\right)+\left(\dfrac{1}{3}+\dfrac{5}{-3}\right)+\dfrac{-3}{8}\)
A=\(2+\dfrac{-4}{3}+\dfrac{-3}{8}\)
A=\(\dfrac{7}{24}\)
B=\(\left(\dfrac{3}{17}+\dfrac{14}{17}\right)+\left(\dfrac{-18}{35}+\dfrac{17}{-35}\right)+\left(\dfrac{-5}{13}+\dfrac{-8}{13}\right)\)
B=\(\dfrac{17}{17}+\dfrac{-35}{35}+\dfrac{-13}{13}\)
B=\(1+\left(-1\right)+\left(-1\right)=-1\)
C=\(\dfrac{-3}{17}+\left(\dfrac{2}{3}+\dfrac{3}{17}\right)\)
C=\(\dfrac{-3}{17}+\dfrac{2}{3}+\dfrac{3}{17}=\left(\dfrac{-3}{17}+\dfrac{3}{17}\right)+\dfrac{2}{3}\)
C=0+\(\dfrac{2}{3}=\dfrac{2}{3}\)
D=\(\left(\dfrac{-1}{6}+\dfrac{5}{-12}\right)+\dfrac{7}{12}\)
D=\(\dfrac{-1}{6}+\dfrac{5}{-12}+\dfrac{7}{12}\)
D=\(\dfrac{-2}{12}+\dfrac{-5}{12}+\dfrac{7}{12}=\left(\dfrac{-2}{12}+\dfrac{-5}{12}\right)+\dfrac{7}{12}\)
D=\(\dfrac{-7}{12}+\dfrac{7}{12}=0\)
các bạn ơi giúp mìh với mìh đag cần gấp ai nhanh và đúng thì mih tick cho
\(\dfrac{2}{1.3}+\dfrac{2}{3.5}+...+\dfrac{2}{2019.2021}\)
= \(2.\left(\dfrac{2}{1.3}+\dfrac{2}{3.5}+...+\dfrac{2}{2019.2021}\right)\)
= \(1.\left(\dfrac{1}{1.3}+\dfrac{1}{3.5}+...+\dfrac{1}{2019.2021}\right)\)
= \(1.\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{2019}-\dfrac{1}{2021}\right)\)
= \(1.\left(1-\dfrac{1}{2021}\right)\)
= \(1.\dfrac{2020}{2021}\)
= \(\dfrac{2020}{2021}\)
Câu 1:
a,\(x=\dfrac{1}{4}+\dfrac{2}{13}\)
\(x=\dfrac{13}{52}+\dfrac{8}{52}=\dfrac{21}{52}\)
Câu 2:
a,\(\dfrac{-2}{5}+\dfrac{3}{-4}+\dfrac{6}{7}+\dfrac{3}{4}+\dfrac{2}{5}\)
\(=\left(\dfrac{-2}{5}+\dfrac{2}{5}\right)+\left(\dfrac{3}{-4}+\dfrac{3}{4}\right)+\dfrac{6}{7}\)
=\(0+0+\dfrac{6}{7}=\dfrac{6}{7}\)
b,\(\dfrac{7}{15}+\dfrac{4}{-9}+\dfrac{-2}{11}+\dfrac{8}{15}+\dfrac{-5}{9}\)
=\(\left(\dfrac{7}{15}+\dfrac{8}{15}\right)+\left(\dfrac{4}{-9}+\dfrac{-5}{9}\right)+\dfrac{-2}{11}\)
=\(\dfrac{15}{15}+\dfrac{-9}{9}+\dfrac{-2}{11}=1+\left(-1\right)+\dfrac{-2}{11}\)
=\(0+\dfrac{-2}{11}=\dfrac{-2}{11}\)
c, \(\dfrac{-5}{7}+\dfrac{5}{13}+\dfrac{-20}{41}+\dfrac{8}{13}+\dfrac{-21}{41}\)
=\(\left(\dfrac{5}{13}+\dfrac{8}{13}\right)+\left(\dfrac{-20}{41}+\dfrac{-21}{41}\right)+\dfrac{-5}{7}\)
=\(\dfrac{13}{13}+\dfrac{-41}{41}+\dfrac{-5}{7}=1+\left(-1\right)+\dfrac{-5}{7}\)
=\(0+\dfrac{-5}{7}=\dfrac{-5}{7}\)
Đây này má Ran mori
a) \(\left(5\dfrac{1}{7}-3\dfrac{3}{11}\right)-2\dfrac{1}{7}-1\dfrac{8}{11}\)
\(=5+\dfrac{1}{7}-3-\dfrac{3}{11}-2-\dfrac{1}{7}-1-\dfrac{8}{11}\)
\(=\left(5-3-2-1\right)+\left(\dfrac{1}{7}-\dfrac{3}{11}-\dfrac{1}{7}-\dfrac{8}{11}\right)\)
\(=-1+\left(\dfrac{1}{7}-\dfrac{1}{7}\right)-\left(\dfrac{3}{11}+\dfrac{8}{11}\right)\)
\(=-1+0-1=-2\)
a)\(\left(5\dfrac{1}{7}-3\dfrac{3}{11}\right)-2\dfrac{1}{7}-1\dfrac{8}{11}\)
= \(\left(5+\dfrac{1}{7}-3+\dfrac{3}{11}\right)-2+\dfrac{1}{7}-1+\dfrac{8}{11}\)
= \(5-\dfrac{1}{7}+3-\dfrac{3}{11}-2+\dfrac{1}{7}-1+\dfrac{8}{11}\)
= \(\left(5-3-2-1\right)+\dfrac{1}{7}+\dfrac{1}{7}+\dfrac{8}{11}-\dfrac{3}{11}\)
= \(-1+2+\dfrac{5}{11}\)
= \(1+\dfrac{5}{11}=\dfrac{1}{1}+\dfrac{5}{11}=\dfrac{11}{11}+\dfrac{5}{11}=\dfrac{16}{11}\)
Vậy :câu a) = \(\dfrac{16}{11}\)
a)
\(3\dfrac{14}{19}+\dfrac{13}{17}+\dfrac{35}{43}+6\dfrac{5}{19}+\dfrac{8}{43}\\ =\left(3\dfrac{14}{19}+6\dfrac{5}{19}\right)+\left(\dfrac{35}{43}+\dfrac{8}{43}\right)+\dfrac{13}{17}\\ =10+1+\dfrac{13}{17}\\ =11\dfrac{13}{17}\)
b)
\(\dfrac{-5}{7}\cdot\dfrac{2}{11}+\dfrac{-5}{7}\cdot\dfrac{9}{11}+1\dfrac{5}{7}\\ =\dfrac{-5}{7}\cdot\left(\dfrac{2}{11}+\dfrac{9}{11}\right)+1\dfrac{5}{7}\\ =\dfrac{-5}{7}\cdot1+1\dfrac{5}{7}\\ =\dfrac{-5}{7}+1\dfrac{5}{7}\\ =1\)
a) \(3\dfrac{14}{19}+\dfrac{13}{17}+\dfrac{35}{43}+6\dfrac{5}{19}+\dfrac{8}{43}\)
\(=\left(3\dfrac{14}{19}+6\dfrac{5}{19}\right)+\left(\dfrac{35}{43}+\dfrac{8}{43}\right)+\dfrac{13}{17}\)
\(=\left[\left(3+6\right)+\left(\dfrac{14}{19}+\dfrac{5}{19}\right)\right]+1+\dfrac{13}{17}\)
\(=\left[9+1\right]+1+\dfrac{13}{17}\)
\(=10+1+\dfrac{13}{17}\)
\(=11+\dfrac{13}{17}\)
\(=\dfrac{187}{17}+\dfrac{13}{17}\)
\(=\dfrac{200}{17}\)
b) \(\dfrac{-5}{7}.\dfrac{2}{11}+\dfrac{-5}{7}.\dfrac{9}{11}+1\dfrac{5}{7}\)
\(=\dfrac{-5}{7}.\left(\dfrac{2}{11}+\dfrac{9}{11}\right)+\dfrac{12}{7}\)
\(=\dfrac{-5}{7}.1+\dfrac{12}{7}\)
\(=\dfrac{-5}{7}+\dfrac{12}{7}\)
\(=\dfrac{7}{7}\)
\(=1\)
c) \(11\dfrac{3}{13}-\left(2\dfrac{4}{7}+5\dfrac{3}{13}\right)\)
= \(11\dfrac{3}{13}-2\dfrac{4}{7}-5\dfrac{3}{13}\)
\(=\left(11\dfrac{3}{13}-5\dfrac{3}{13}\right)-2\dfrac{4}{7}\)
\(=\left[\left(11-5\right)+\left(\dfrac{3}{13}-\dfrac{3}{13}\right)\right]-\dfrac{18}{7}\)
\(=\left[6+0\right]-\dfrac{18}{7}\)
\(=6-\dfrac{18}{7}\)
\(=\dfrac{42}{7}-\dfrac{18}{7}\)
\(=\dfrac{24}{7}\)
d) \(\dfrac{2}{7}.5\dfrac{1}{4}-\dfrac{2}{7}.3\dfrac{1}{4}\)
\(=\dfrac{2}{7}.\left(5\dfrac{1}{4}-3\dfrac{1}{4}\right)\)
\(=\dfrac{2}{7}.\left[\left(5-3\right)+\left(\dfrac{1}{4}-\dfrac{1}{4}\right)\right]\)
\(=\dfrac{2}{7}.\left[2+0\right]\)
\(=\dfrac{2}{7}.2\)
= \(\dfrac{4}{7}\)
\(\dfrac{2}{1\cdot2}+\dfrac{2}{2\cdot3}+\dfrac{2}{3\cdot4}+...+\dfrac{2}{19\cdot20}\)
\(=2\cdot\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{19}-\dfrac{1}{20}\right)\)
\(=2\cdot\left(1-\dfrac{1}{20}\right)\)
\(=2\cdot\dfrac{19}{20}\)
\(=\dfrac{19}{10}\)