\(\dfrac{2}{15}-\dfrac{7}{1}+\dfrac{13}{15}+\dfrac{1}{2}\)
1, \(\dfrac{-5}{7}\) . \(\dfrac{2}{11}\) + \(\dfrac{-5}{7}\) . \(\dfrac{9}{11}\) + \(1\dfrac{5}{7}\)
2,\(-3\dfrac{4}{13}\) . \(15\dfrac{3}{41}\) + \(3\dfrac{4}{13}\) . \(2\dfrac{3}{41}\)
3, \(\dfrac{4}{5}\) . \(15\dfrac{1}{4}\) - \(\dfrac{4}{5}\) . \(15\dfrac{1}{3}\) + \(\dfrac{11}{30}\)
4,\(\dfrac{4}{20}\) + \(\dfrac{16}{42}\) + \(\dfrac{6}{15}\) - \(\dfrac{3}{5}\) + \(\dfrac{2}{21}\) - \(\dfrac{10}{21}\) + \(\dfrac{3}{10}\)
Giúp mik nha. Cảm ơn
\(\dfrac{2}{15}-\dfrac{7}{2}+\dfrac{13}{15}+\dfrac{1}{2}\)
= \(\left(\dfrac{2}{15}+\dfrac{13}{15}\right)\)- \(\left(\dfrac{7}{2}-\dfrac{1}{2}\right)\)
= \(\dfrac{15}{15}-\dfrac{6}{2}\)
= 1 -3
= -2
tính
a)\(\dfrac{-10}{11}.\dfrac{8}{9}+\dfrac{7}{18}.\dfrac{10}{11}\)
b)\(\dfrac{3}{14}:\dfrac{1}{28}-\dfrac{13}{21}:\dfrac{1}{28}+\dfrac{29}{42}:\dfrac{1}{28}-8\)
c)\(-1\dfrac{5}{7}.15+\dfrac{2}{7}\left(-15\right)+\left(-105\right).\left(\dfrac{2}{3}-\dfrac{4}{5}+\dfrac{1}{7}\right)\)
a)\(\dfrac{-10}{11}.\dfrac{8}{9}+\dfrac{7}{18}.\dfrac{10}{11}\)
=\(\dfrac{10}{11}.\dfrac{-8}{9}+\dfrac{7}{18}.\dfrac{10}{11}\)
=\(\dfrac{10}{11}(\dfrac{-8}{9}+\dfrac{7}{18})\)
=\(\dfrac{10}{11}.\dfrac{-1}{2}\)
=\(\dfrac{-5}{11}\)
b;
B = \(\dfrac{3}{14}\) : \(\dfrac{1}{28}\) - \(\dfrac{13}{21}\): \(\dfrac{1}{28}\) + \(\dfrac{29}{42}\) : \(\dfrac{1}{28}\) - 8
B = (\(\dfrac{3}{14}\) - \(\dfrac{13}{21}\) + \(\dfrac{29}{42}\)) - 8
B = (\(\dfrac{9}{42}\) - \(\dfrac{26}{42}\) + \(\dfrac{29}{42}\)) - 8
B = (\(\dfrac{-17}{42}\) + \(\dfrac{29}{42}\)) - 8
B = \(\dfrac{2}{7}\) - 8
B = \(\dfrac{2}{7}-\dfrac{56}{7}\)
B = - \(\dfrac{54}{7}\)
c; C = -1\(\dfrac{5}{7}\).15 + \(\dfrac{2}{7}\)(-15) + (-105).(\(\dfrac{2}{3}\) - \(\dfrac{4}{5}\) + \(\dfrac{1}{7}\))
C = - 15.(- 1 - \(\dfrac{5}{7}\) + \(\dfrac{2}{7}\) + \(\dfrac{14}{3}\) - \(\dfrac{28}{5}\) + \(1\))
C = -15.[(1 - 1) - (\(\dfrac{5}{7}\) - \(\dfrac{2}{7}\)) + \(\dfrac{14}{3}\) - \(\dfrac{28}{5}\)]
C = -15.[0 - \(\dfrac{3}{7}\) + \(\dfrac{14}{3}\) - \(\dfrac{28}{5}\)]
C = -15 . [- \(\dfrac{45}{105}\) + \(\dfrac{490}{105}\) - \(\dfrac{588}{105}\)]
C = -15. [ \(\dfrac{445}{105}\) - \(\dfrac{588}{105}\)]
C = - 15.(- \(\dfrac{143}{105}\))
C = \(\dfrac{143}{7}\)
Bài 1: Thực hiện phép tính:
1) \(\dfrac{-17}{30}-\dfrac{11}{-15}+\dfrac{-7}{12}\)
2) \(\dfrac{-5}{9}+\dfrac{5}{9}:\left(1\dfrac{2}{3}-2\dfrac{1}{12}\right)\)
3) \(\dfrac{-7}{25}.\dfrac{11}{13}+\dfrac{-7}{25}.\dfrac{2}{13}-\dfrac{18}{25}\)
1) âm năm phần 12
2) âm mười bảy phần 9
3) -1
Đây là đáp án còn làm bài từ làm nhé
BT2: Tính nhanh
15) \(\left(\dfrac{1}{2}-\dfrac{13}{14}\right):\dfrac{5}{7}-\left(-\dfrac{2}{21}+\dfrac{1}{7}\right):\dfrac{5}{7}\)
16)\(\dfrac{3}{5}:\left(\dfrac{-1}{15}-\dfrac{1}{6}\right)+\dfrac{3}{5}:\left(\dfrac{-1}{3}-1\dfrac{1}{15}\right)\)
15: \(=\dfrac{7-13}{14}\cdot\dfrac{7}{5}-\dfrac{-2+3}{21}\cdot\dfrac{7}{5}\)
\(=\dfrac{7}{5}\left(-\dfrac{6}{14}-\dfrac{1}{21}\right)\)
\(=\dfrac{7}{5}\cdot\dfrac{-10}{21}=\dfrac{-2}{3}\)
16: \(=\dfrac{3}{5}:\dfrac{-2-5}{30}+\dfrac{3}{5}:\left(\dfrac{-1}{3}-\dfrac{16}{15}\right)\)
\(=\dfrac{3}{5}\cdot\dfrac{-30}{7}+\dfrac{3}{5}:\dfrac{-5-16}{15}\)
\(=\dfrac{3}{5}\cdot\dfrac{-30}{7}+\dfrac{3}{5}\cdot\dfrac{-5}{7}\)
\(=\dfrac{3}{5}\cdot\left(-5\right)=-3\)
Tính rồi rút gọn (theo mẫu):
Mẫu: \(\dfrac{9}{10}-\dfrac{4}{10}=\dfrac{9-4}{10}=\dfrac{5}{10}=\dfrac{1}{2}\) |
a) \(\dfrac{15}{8}-\dfrac{13}{8}\) b) \(\dfrac{7}{15}-\dfrac{2}{15}\) c) \(\dfrac{11}{12}-\dfrac{2}{12}\) d) \(\dfrac{19}{7}-\dfrac{5}{7}\)
a: \(\dfrac{15}{8}-\dfrac{13}{8}=\dfrac{15-13}{8}=\dfrac{2}{8}=\dfrac{1}{4}\)
b: \(\dfrac{7}{15}-\dfrac{2}{15}=\dfrac{7-2}{15}=\dfrac{5}{15}=\dfrac{1}{3}\)
c: \(\dfrac{11}{12}-\dfrac{2}{12}=\dfrac{11-2}{12}=\dfrac{9}{12}=\dfrac{3}{4}\)
d: \(\dfrac{19}{7}-\dfrac{5}{7}=\dfrac{19-5}{7}=\dfrac{14}{7}=2\)
1.Thực hiện phép tính
a) \(\dfrac{4}{5}\) + \(\dfrac{6}{5}\) : \(\dfrac{7}{15}\) - \(\dfrac{3}{7}\)
b) \(\dfrac{9}{13}\) . \(\dfrac{5}{3}\) + \(\dfrac{-6}{13}\) .\(\dfrac{9}{13}\) - \(\dfrac{9}{13}\) . \(\dfrac{20}{13}\)
2.Tìm x,biết:
a) \(\dfrac{2}{5}\) . x - \(\dfrac{1}{3}\) = 1\(\dfrac{2}{3}\)
b) \(\dfrac{4}{9}\) - \(\dfrac{5}{9}\) : x = \(\dfrac{7}{3}\)
c) \(\dfrac{4}{5}\) . x + \(\dfrac{-7}{3}\) . x = \(\dfrac{\left(-2\right)^3}{5}\)
\(\dfrac{99}{100}:\left(\dfrac{1}{4}-\dfrac{1}{12}+\dfrac{1}{3}\right)-\left(\dfrac{-7}{5}\right)^2\)
\(\dfrac{13}{15}\cdot0,25\cdot3+\left(\dfrac{8}{15}-1\dfrac{19}{60}\right):1\dfrac{23}{24}\)
a: \(=\dfrac{99}{100}:\left(\dfrac{3}{12}-\dfrac{1}{12}+\dfrac{4}{12}\right)-\dfrac{49}{25}\)
\(=\dfrac{99}{100}:\dfrac{1}{2}-\dfrac{49}{25}\)
\(=\dfrac{99}{50}-\dfrac{98}{50}=\dfrac{1}{50}\)
b: \(=\dfrac{13}{15}\cdot\dfrac{1}{4}\cdot3+\left(\dfrac{32}{60}-1-\dfrac{19}{60}\right):\dfrac{47}{24}\)
\(=\dfrac{39}{60}+\dfrac{-19}{60}\cdot\dfrac{24}{47}\)
=459/940