=\(\dfrac{2}{3}\times\left(\dfrac{4}{9}+\dfrac{5}{9}\right)=\dfrac{2}{3}\times\dfrac{9}{9}=\dfrac{2}{3}\)

\(=\dfrac{2}{3}\times\left(\dfrac{4}{9}+\dfrac{5}{9}\right)\\ =\dfrac{2}{3}\times1=\dfrac{2}{3}\)

\(=\dfrac{2}{3}\left(\dfrac{4}{9}+\dfrac{5}{9}\right)=\dfrac{2}{3}\)

2/5

2/5

= 2/5 x ( 8/7 - 1/7)

= 2/5 x 1

= 2/5

=2+8/7-2/7

=2+6/7

=20/7

14,5 x  2,7 + 2,7  x 5,5 - 2,7 x  2  x 5.

= 2,7 x 14,5 + 2,7 x 5,5 - 2,7 x 10

= 2,7 x ( 14,5 + 5,5 - 10 )

= 2,7 x 10

= 27

\(\dfrac{2}{9}+\dfrac{3}{7}+\dfrac{15}{13}+\dfrac{4}{7}-\dfrac{2}{13}+\dfrac{7}{9}\)

\(=\left(\dfrac{2}{9}+\dfrac{7}{9}\right)+\left(\dfrac{3}{7}+\dfrac{4}{7}\right)+\left(\dfrac{15}{13}-\dfrac{2}{13}\right)\)

\(=\dfrac{9}{9}+\dfrac{7}{7}+\dfrac{13}{13}=1+1+1=3\)

`2/9+3/7+15/13+4/7-2/13+7/9`

`= (2/9 + 7/9 ) + ( 3/7 + 4/7 ) + ( 15/13-2/13 ) `

`= 9/9 + 7/7 + 13/13`

`=   1 + 1 + 1 = 3`

2/2/19+1/5-5/3/19+(-5/3/19-6/5)

=2/2/19+(1/5-6/5)+(5/3/19-5/3/19)

=19-1+0=18

\(=\dfrac{2}{57}+\dfrac{1}{5}-\dfrac{5}{57}-\dfrac{5}{57}-\dfrac{6}{5}=-1-\dfrac{8}{57}=-\dfrac{65}{57}\)

a: \(\left(-\dfrac{5}{6}+\dfrac{2}{5}\right):\dfrac{3}{8}+\left(\dfrac{4}{5}-\dfrac{11}{30}\right):\dfrac{3}{8}\)

\(=\left(-\dfrac{5}{6}+\dfrac{2}{5}\right)\cdot\dfrac{8}{3}+\left(\dfrac{4}{5}-\dfrac{11}{30}\right)\cdot\dfrac{8}{3}\)

\(=\dfrac{8}{3}\left(-\dfrac{5}{6}+\dfrac{2}{5}+\dfrac{4}{5}-\dfrac{11}{30}\right)\)

\(=\dfrac{8}{3}\cdot\dfrac{-25+36-11}{30}\)

=0

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

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

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

\(=\dfrac{7}{3}\cdot0=0\)

c: \(\dfrac{-13}{18}\cdot\dfrac{5}{8}+\dfrac{-5}{18}\cdot\dfrac{2}{9}+\dfrac{-13}{18}\cdot\dfrac{3}{8}+\dfrac{-5}{18}\cdot\dfrac{7}{9}\)

\(=\left(-\dfrac{13}{18}\cdot\dfrac{5}{8}+\dfrac{-13}{18}\cdot\dfrac{3}{8}\right)+\left(-\dfrac{5}{18}\cdot\dfrac{2}{9}+\dfrac{-5}{18}\cdot\dfrac{7}{9}\right)\)

\(=-\dfrac{13}{18}\left(\dfrac{5}{8}+\dfrac{3}{8}\right)+\dfrac{-5}{18}\left(\dfrac{2}{9}+\dfrac{7}{9}\right)\)

\(=-\dfrac{13}{18}-\dfrac{5}{18}=-\dfrac{18}{18}=-1\)

d: Sửa đề: \(\dfrac{-11}{19}\cdot\dfrac{4}{9}+\dfrac{-8}{19}\cdot\dfrac{3}{7}+\dfrac{-11}{19}\cdot\dfrac{5}{9}+\dfrac{-8}{19}\cdot\dfrac{4}{7}\)

\(=\left(-\dfrac{11}{19}\cdot\dfrac{4}{9}+\dfrac{-11}{19}\cdot\dfrac{5}{9}\right)+\left(\dfrac{-8}{19}\cdot\dfrac{3}{7}+\dfrac{-8}{19}\cdot\dfrac{4}{7}\right)\)

\(=-\dfrac{11}{19}\left(\dfrac{4}{9}+\dfrac{5}{9}\right)+\dfrac{-8}{19}\left(\dfrac{3}{7}+\dfrac{4}{7}\right)\)

\(=-\dfrac{11}{19}-\dfrac{8}{19}=-\dfrac{19}{19}=-1\)

\(a.\left(-\dfrac{5}{6}+\dfrac{2}{5}\right):\dfrac{3}{8}+\left(\dfrac{4}{5}-\dfrac{11}{30}\right):\dfrac{3}{8}\)

\(=\left(-\dfrac{13}{30}\right):\dfrac{3}{8}+\dfrac{13}{30}:\dfrac{3}{8}\)

\(=\left[\left(-\dfrac{13}{30}+\dfrac{13}{30}\right)\right]:\dfrac{3}{8}\)

\(=0:\dfrac{3}{8}=0\)

\(b.\left(-\dfrac{3}{4}+\dfrac{2}{5}\right):\dfrac{3}{7}+\left(\dfrac{3}{5}+-\dfrac{1}{4}\right):\dfrac{3}{7}\)

\(=\left(-\dfrac{7}{20}\right):\dfrac{3}{7}+\dfrac{7}{20}:\dfrac{3}{7}\)

\(=\left[\left(-\dfrac{7}{20}+\dfrac{7}{20}\right)\right]:\dfrac{3}{7}=0:\dfrac{3}{7}=0\)

\(c.-\dfrac{13}{18}.\dfrac{5}{8}+-\dfrac{5}{18}.\dfrac{2}{9}+-\dfrac{13}{18}.\dfrac{3}{8}+-\dfrac{5}{18}.\dfrac{7}{9}\)

\(=\left(\dfrac{5}{8}+\dfrac{3}{8}\right).-\dfrac{13}{18}+\left(\dfrac{2}{9}+\dfrac{7}{9}\right).-\dfrac{5}{18}\)

\(=1.-\dfrac{13}{18}+1.-\dfrac{5}{18}=-\dfrac{13}{18}+-\dfrac{5}{18}=-1\)

\(d.-\dfrac{11}{19}.\dfrac{4}{9}+\dfrac{-8}{19}.\dfrac{3}{7}+-\dfrac{11}{19}.\dfrac{5}{9}+-\dfrac{9}{19}.\dfrac{4}{7}\)

\(=\left(\dfrac{4}{9}+\dfrac{5}{9}\right).-\dfrac{11}{19}+-\dfrac{24}{133}+-\dfrac{36}{133}\)

\(=-\dfrac{11}{19}+-\dfrac{60}{133}=-\dfrac{137}{133}\)

\(\dfrac{2}{15}\times\dfrac{1}{8}\times\dfrac{15}{12}=1\times\dfrac{1}{8}\times\dfrac{1}{6}=\dfrac{1}{48}\)

=(2/15x15/12)x1/8

= 1/6x 1/8

= 1/48

( nếu có sai thì thui nha)

1/48

\(\dfrac{4}{5}\)+\(\dfrac{-5}{4}\)=\(\dfrac{0}{4}\)

\(\dfrac{-1}{3}\)+\(\dfrac{2}{5}-\dfrac{5}{6}\)=\(\dfrac{-10}{30}+\dfrac{12}{30}-\dfrac{25}{30}\)=\(\dfrac{-23}{30}\)

\(\dfrac{2}{3}-\dfrac{5}{7}.\dfrac{14}{25}\)=\(\dfrac{2}{3}-\dfrac{2}{5}\)=\(\dfrac{10}{15}-\dfrac{6}{15}\)=\(\dfrac{4}{15}\)

a: \(=\dfrac{16-25}{20}=\dfrac{-9}{20}\)

b: \(=\dfrac{-10}{30}+\dfrac{12}{30}-\dfrac{25}{30}=-\dfrac{23}{30}\)

c: \(=\dfrac{2}{3}-\dfrac{1}{5}\cdot2=\dfrac{2}{3}-\dfrac{2}{5}=\dfrac{10-6}{15}=\dfrac{4}{15}\)

d: \(=\dfrac{12-25}{30}\cdot\dfrac{15}{8}-1=\dfrac{-13}{16}-1=\dfrac{-29}{16}\)

e: \(=\dfrac{-3}{13}-\dfrac{10}{13}-\dfrac{7}{17}+\dfrac{24}{17}-\dfrac{5}{19}-\dfrac{14}{19}=-1\)

- Viết phép tính sao cho các số cùng hàng thẳng cột với nhau.

- Tính trừ lần lượt từ phải sang trái.