A -\(\dfrac{24}{25}\)

B -\(\dfrac{5}{21}\)

C -\(\dfrac{24}{47}\)

D -\(\dfrac{19}{42}\)

tick cho mk

a) \(\dfrac{-5}{9}+\dfrac{8}{15}+\dfrac{-2}{11}+\dfrac{4}{-9}+\dfrac{7}{15}\)

=\(\left(\dfrac{-5}{9}+\dfrac{-4}{9}\right)+\left(\dfrac{8}{15}+\dfrac{7}{15}\right)+\dfrac{-2}{11}\)

=\(\left(-1\right)+1+\dfrac{-2}{11}\)

=\(\dfrac{-2}{11}\)

b) \(\left(\dfrac{-5}{12}+\dfrac{6}{11}\right)+\left(\dfrac{7}{17}+\dfrac{5}{11}+\dfrac{5}{12}\right)\)

=\(\dfrac{-5}{12}+\dfrac{6}{11}+\dfrac{7}{17}+\dfrac{5}{11}+\dfrac{5}{12}\)

=\(\left(\dfrac{-5}{12}+\dfrac{5}{12}\right)+\left(\dfrac{6}{11}+\dfrac{5}{11}\right)+\dfrac{7}{17}\)

=\(0+0+\dfrac{7}{17}\)

=\(\dfrac{7}{17}\)

c) A= \(49\dfrac{8}{23}-\left(5\dfrac{7}{32}+14\dfrac{8}{23}\right)\)

A=\(49\dfrac{8}{23}-5\dfrac{7}{32}-14\dfrac{8}{23}\)

A=\(\left(49\dfrac{8}{23}-14\dfrac{8}{23}\right)-5\dfrac{7}{32}\)

A=\(35-5\dfrac{7}{32}\)

A=\(35-\dfrac{167}{32}=\dfrac{953}{32}\)

d) C=\(\dfrac{-3}{7}.\dfrac{5}{9}+\dfrac{4}{9}.\dfrac{-3}{7}+2\dfrac{3}{7}\)

C=\(\dfrac{-3}{7}.\left(\dfrac{5}{9}+\dfrac{4}{9}\right)+\dfrac{17}{7}\)

C=\(\dfrac{-3}{7}.1+\dfrac{17}{7}\)

C=\(\dfrac{-3}{7}+\dfrac{17}{7}=2\)

a, `(-5)/9+8/15+(-2)/11+4/(-9)+7/15`

`=-5/9+8/15-2/11-4/9+7/15`

`=(-5/9-4/9)+(8/15+7/15)-2/11`

`=-9/9+15/15-2/11`

`=-1+1-2/11`

`=-2/11`

b, `((-5)/12+6/11)+(7/17+5/11+5/12)`

`=-5/12+6/11+7/17+5/11+5/12`

`=(-5/12+5/12)+(6/11+5/11)+7/17`

`=0+11/11+7/17`

`=1+7/17`

`=17/17+7/17`

`=24/17`

c, `A=49 8/23 - (5 7/32 + 14 8/23)`

`A=49 8/23 - 5 7/32 - 14 8/23`

`A=(49 8/23 - 14 8/23)-5 7/32`

`A=35 - 167/32`

`A=953/32`

d, `C=(-3)/7.5/9+4/9.(-3)/7+2 3/7`

`C=-3/7 . 5/9-4/9 . 3/7+17/7`

`C=-3/7.(5/9+4/9)+17/7`

`C=-3/7 . 1+17/7`

`C=2`

a) Ta có: \(\dfrac{-5}{9}+\dfrac{8}{15}+\dfrac{-2}{11}+\dfrac{-4}{9}+\dfrac{7}{15}\)

\(=\left(\dfrac{-5}{9}+\dfrac{-4}{9}\right)+\left(\dfrac{8}{15}+\dfrac{7}{15}\right)-\dfrac{2}{11}\)

\(=-1+1-\dfrac{2}{11}\)

\(=-\dfrac{2}{11}\)

b) Ta có: \(\left(\dfrac{-5}{12}+\dfrac{6}{11}\right)+\left(\dfrac{7}{11}+\dfrac{5}{11}+\dfrac{5}{12}\right)\)

\(=\left(\dfrac{-5}{12}+\dfrac{5}{12}\right)+\left(\dfrac{6}{11}+\dfrac{7}{11}+\dfrac{5}{11}\right)\)

\(=\dfrac{18}{11}\)

c) Ta có: \(A=49\dfrac{8}{23}-\left(5\dfrac{7}{32}+14\dfrac{8}{23}\right)\)

\(=49+\dfrac{8}{23}-14-\dfrac{8}{23}-5-\dfrac{7}{32}\)

\(=30-\dfrac{7}{32}=\dfrac{953}{32}\)

1) \(\left(-1\dfrac{1}{5}+x\right):\left(-3\dfrac{3}{5}\right)=\dfrac{-7}{4}+\dfrac{1}{4}:\dfrac{1}{8}\)

\(\Leftrightarrow\left(-1\dfrac{1}{5}+x\right):\left(-3\dfrac{3}{5}\right)=\dfrac{-7}{4}+2\)

\(\Leftrightarrow\left(-1\dfrac{1}{5}+x\right):\left(-3\dfrac{3}{5}\right)=\dfrac{1}{4}\)

\(\Leftrightarrow-1\dfrac{1}{5}+x=\dfrac{1}{4}.\left(-3\dfrac{3}{5}\right)\)

\(\Leftrightarrow-1\dfrac{1}{5}+x=\dfrac{1}{4}.\left(-\dfrac{18}{5}\right)\)

\(\Leftrightarrow-1\dfrac{1}{5}+x=-\dfrac{9}{10}\)

\(\Leftrightarrow x=\left(-\dfrac{9}{10}\right)-\left(-1\dfrac{1}{5}\right)\)

\(\Leftrightarrow x=\dfrac{3}{10}\)

\(\dfrac{11}{8}\cdot\left[\left(-\dfrac{5}{11}:\dfrac{13}{8}-\dfrac{5}{11}:\dfrac{13}{15}\right)+\dfrac{-6}{33}\right]+\dfrac{-3}{4}\)

\(=\dfrac{11}{8}\cdot\left[\left(-\dfrac{5}{11}\cdot\dfrac{8}{13}-\dfrac{5}{11}\cdot\dfrac{15}{13}\right)-\dfrac{2}{11}\right]-\dfrac{3}{4}\)

\(=\dfrac{11}{8}\cdot\left[-\dfrac{5}{11}\cdot\left(\dfrac{8}{13}+\dfrac{15}{13}\right)-\dfrac{2}{11}\right]-\dfrac{3}{4}\)

\(=\dfrac{11}{8}\cdot\left(-\dfrac{5}{11}\cdot\dfrac{23}{13}-\dfrac{2}{11}\right)-\dfrac{3}{4}\)

\(=\dfrac{11}{8}\cdot\left(-\dfrac{115}{143}-\dfrac{2}{11}\right)-\dfrac{3}{4}\)

\(=\dfrac{11}{8}\cdot\dfrac{-141}{143}-\dfrac{3}{4}\)

\(=-\dfrac{141}{104}-\dfrac{3}{4}\)

\(=-\dfrac{219}{104}\)

chắc h có mấy thành cay r nên ko làm bn lên mạng tải phẩn mêm có cánh iair đó :D

\(3-1\dfrac{1}{2}-x+\dfrac{5}{4}=2-\left|1\dfrac{1}{8}-\dfrac{5}{12}\right|\)

\(3-1\dfrac{1}{2}-x+\dfrac{5}{4}=2-\dfrac{17}{24}\)

\(x=3-1\dfrac{1}{2}+\dfrac{5}{4}-2+\dfrac{17}{24}\)

\(x=\dfrac{35}{24}\)

bn làm từng bước nhé :D

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 = \(\dfrac{143}{7}\)

\(a.-8:\left(4\dfrac{1}{5}x+\dfrac{3}{10}\right)=4\dfrac{4}{9}\)

\(4\dfrac{1}{5}x+\dfrac{3}{10}=\left(-8\right):4\dfrac{4}{9}\)

\(4\dfrac{1}{5}x+\dfrac{3}{10}=\dfrac{-9}{5}\)

\(4\dfrac{1}{5}x=\dfrac{-9}{5}-\dfrac{3}{10}\)

\(4\dfrac{1}{5}x=\dfrac{-21}{10}\)

\(x=\dfrac{-21}{10}:\dfrac{21}{5}\)

\(x=\dfrac{-1}{2}\)

Vay \(x=\dfrac{-1}{2}\).

\(b.4\dfrac{2}{3}-\left(\dfrac{3}{5}:x\right)=-20\%\)

\(\dfrac{14}{3}-\left(\dfrac{3}{5}:x\right)=\dfrac{-1}{5}\)

\(\dfrac{3}{5}:x=\dfrac{14}{3}-\dfrac{-1}{5}\)

\(\dfrac{3}{5}:x=\dfrac{73}{15}\)

\(x=\dfrac{3}{5}:\dfrac{73}{15}\)

\(x=\dfrac{9}{73}\)

Vay \(x=\dfrac{9}{73}\).

Câu c; d; e tương tự nhé.