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10 tháng 4 2017

\(\dfrac{\dfrac{2}{3}+\dfrac{2}{7}-\dfrac{1}{14}}{-1-\dfrac{3}{7}+\dfrac{3}{28}}\)

\(=\dfrac{\dfrac{2}{3}+\dfrac{2}{7}-\dfrac{2}{28}}{-\dfrac{3}{3}-\dfrac{3}{7}+\dfrac{3}{28}}\\ =\dfrac{2\left(\dfrac{1}{3}+\dfrac{1}{7}-\dfrac{1}{28}\right)}{-3\left(\dfrac{1}{3}+\dfrac{1}{7}-\dfrac{1}{28}\right)}\\ =\dfrac{2}{-3}\\ =-\dfrac{2}{3}\)

a)\(\dfrac{-10}{11}.\dfrac{8}{9}+\dfrac{7}{18}.\dfrac{10}{11}\)

=\(\dfrac{10}{11}.\dfrac{-8}{9}+\dfrac{7}{18}.\dfrac{10}{11}\)

=\(\dfrac{10}{11}(\dfrac{-8}{9}+\dfrac{7}{18})\)

=\(\dfrac{10}{11}.\dfrac{-1}{2}\)

=\(\dfrac{-5}{11}\)

7 tháng 8

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 = - \(\dfrac{54}{7}\)

22 tháng 7 2021

a) `6/18+ (-14)/21 = 1/3 - 2/3=-1/3`

b) `3/5 - (-1/2) = 3/5+1/2=11/10`

c) `2/7 : 3/4 = 2/7 . 4/3 = 8/21`

d) `(-28)/33 . (-3)/4 = 28/33 . 3/4=7/11`

a) \(\dfrac{6}{18}+\dfrac{-14}{21}=\dfrac{1}{3}-\dfrac{2}{3}=\dfrac{-1}{3}\)

b) \(\dfrac{3}{5}-\dfrac{-1}{2}=\dfrac{3}{5}+\dfrac{1}{2}=\dfrac{6}{10}+\dfrac{5}{10}=\dfrac{11}{10}\)

c) \(\dfrac{2}{7}:\dfrac{3}{4}=\dfrac{2}{7}\cdot\dfrac{4}{3}=\dfrac{8}{21}\)

d) \(\dfrac{-28}{33}\cdot\dfrac{-3}{4}=\dfrac{28}{4}\cdot\dfrac{3}{33}=7\cdot\dfrac{1}{11}=\dfrac{7}{11}\)

a: \(A=\dfrac{-7}{28}\cdot\dfrac{15}{25}=\dfrac{-1}{4}\cdot\dfrac{3}{5}=\dfrac{-3}{20}\)

b: \(B=\dfrac{-5\cdot7}{14\cdot\left(-3\right)}=\dfrac{35}{42}=\dfrac{5}{6}\)

c: \(C=\dfrac{-1}{5}-\dfrac{1}{5}\cdot\dfrac{3}{5}=\dfrac{-1}{5}-\dfrac{3}{25}=\dfrac{-8}{25}\)

d: \(D=\dfrac{-3}{4}-\dfrac{1}{4}=-1\)

e: \(E=\dfrac{-4}{5}\left(1-\dfrac{15}{16}\right)=\dfrac{-4}{5}\cdot\dfrac{1}{16}=\dfrac{-1}{20}\)

f: \(F=\dfrac{6-7}{4}\cdot\dfrac{4+12}{22}=\dfrac{-1}{4}\cdot\dfrac{8}{11}=\dfrac{-2}{11}\)

20 tháng 8 2017

a, \(4\dfrac{5}{37}\)-\(3\dfrac{4}{5}\)+ \(8\dfrac{15}{29}\)- \(3\dfrac{5}{37}\)+ \(6\dfrac{14}{29}\)

=(\(4\dfrac{5}{37}\)-\(3\dfrac{5}{37}\))+(\(8\dfrac{15}{29}\)+\(6\dfrac{14}{29}\))-\(3\dfrac{4}{5}\)

=(4-3)+(\(\dfrac{5}{37}\)-\(\dfrac{5}{37}\))+(8+6)+(\(\dfrac{15}{29}\)+\(\dfrac{14}{29}\))-3\(\dfrac{4}{5}\)

=1+ 15-\(3\dfrac{4}{5}\)=13-\(\dfrac{4}{5}\)=\(\dfrac{61}{5}\)

b, 60\(\dfrac{7}{13}\)+ 50\(\dfrac{8}{13}\)-11\(\dfrac{2}{13}\)

=(60+50-11)+(\(\dfrac{7}{13}\)+ \(\dfrac{8}{13}\)-\(\dfrac{2}{13}\))

=99+1=100

c, đáp án bằng \(\dfrac{-2}{3}\). bạn tự tính nha

19 tháng 12 2020

a) Ta có: \(\left|5\cdot0.6+\dfrac{2}{3}\right|-\dfrac{1}{3}\)

\(=\left|3+\dfrac{2}{3}\right|-\dfrac{1}{3}\)

\(=3+\dfrac{2}{3}-\dfrac{1}{3}\)

\(=3+\dfrac{1}{3}=\dfrac{10}{3}\)

b) Ta có: \(\left(0.25-1\dfrac{1}{4}\right):5-\dfrac{1}{5}\cdot\left(-3\right)^2\)

\(=\left(\dfrac{1}{4}-\dfrac{5}{4}\right)\cdot\dfrac{1}{5}-\dfrac{1}{5}\cdot9\)

\(=\dfrac{-4}{4}\cdot\dfrac{1}{5}-\dfrac{1}{5}\cdot9\)

\(=\dfrac{1}{5}\cdot\left(-1-9\right)\)

\(=-10\cdot\dfrac{1}{5}=-2\)

c) Ta có: \(\dfrac{14}{17}\cdot\dfrac{7}{5}-\dfrac{-3}{17}:\dfrac{5}{7}\)

\(=\dfrac{14}{17}\cdot\dfrac{7}{5}-\dfrac{-3}{17}\cdot\dfrac{7}{5}\)

\(=\dfrac{7}{5}\cdot\left(\dfrac{14}{17}+\dfrac{3}{17}\right)\)

\(=\dfrac{7}{5}\cdot1=\dfrac{7}{5}\)

d) Ta có: \(\dfrac{7}{16}+\dfrac{-9}{25}+\dfrac{9}{16}+\dfrac{-16}{25}\)

\(=\left(\dfrac{7}{16}+\dfrac{9}{16}\right)-\left(\dfrac{9}{25}+\dfrac{16}{25}\right)\)

\(=\dfrac{16}{16}-\dfrac{25}{25}\)

\(=1-1=0\)

e) Ta có: \(\dfrac{5}{6}+2\sqrt{\dfrac{4}{9}}\)

\(=\dfrac{5}{6}+2\cdot\dfrac{2}{3}\)

\(=\dfrac{5}{6}+\dfrac{4}{3}\)

\(=\dfrac{5}{6}+\dfrac{8}{6}=\dfrac{13}{6}\)

4 tháng 4 2021

1,

a, \(\left(\dfrac{-4}{3}+\dfrac{1}{3}\right).\dfrac{5}{12}\)=-\(\dfrac{5}{12}\)

b, \(\dfrac{16}{5}+\left(\dfrac{-45}{14}\right):\dfrac{3}{28}\)

=\(\dfrac{-2}{15}\)

2,

a, 2x+19=25

=>x=3

b, \(-\dfrac{2}{9}x=\dfrac{1}{3}\)

=>x=\(\dfrac{-3}{2}\)

Bài 1: 

a) Ta có: \(\dfrac{-4}{3}\cdot\dfrac{5}{12}+\dfrac{1}{3}\cdot\dfrac{5}{12}\)

\(=\dfrac{5}{12}\cdot\left(\dfrac{-4}{3}+\dfrac{1}{3}\right)\)

\(=\dfrac{-5}{12}\)

b) Ta có: \(3\dfrac{1}{5}+\left(\dfrac{2}{7}-\dfrac{7}{2}\right):\dfrac{3}{28}\)

\(=\dfrac{16}{5}+\left(\dfrac{4}{14}-\dfrac{49}{14}\right):\dfrac{3}{28}\)

\(=\dfrac{16}{5}+\dfrac{-45}{14}\cdot\dfrac{28}{3}\)

\(=\dfrac{16}{5}-30=\dfrac{-134}{5}\)

1: Ta có: \(23\dfrac{1}{4}\cdot\dfrac{7}{5}-13\dfrac{1}{4}:\dfrac{5}{7}\)

\(=\dfrac{93}{4}\cdot\dfrac{7}{5}-\dfrac{53}{4}\cdot\dfrac{7}{5}\)

\(=\dfrac{7}{5}\cdot10=14\)

2: Ta có: \(\left(1+\dfrac{2}{3}-\dfrac{1}{4}\right)\left(\dfrac{4}{5}-\dfrac{3}{4}\right)^2\)

\(=\dfrac{12+8-3}{12}\cdot\dfrac{1}{400}\)

\(=\dfrac{17}{12}\cdot\dfrac{1}{400}=\dfrac{17}{4800}\)

a: \(=\left(\dfrac{28}{42}+\dfrac{12}{42}-\dfrac{3}{42}\right):\left(\dfrac{-28}{28}-\dfrac{12}{28}+\dfrac{3}{28}\right)\)

\(=\dfrac{37}{42}:\dfrac{-37}{28}=\dfrac{-28}{42}=-\dfrac{2}{3}\)

b: \(=\dfrac{2+8+18+32+50}{12+48+108+192+300}=\dfrac{110}{660}=\dfrac{1}{6}\)