1:

a: \(=\dfrac{-4}{7}+\dfrac{4}{7}+\dfrac{3}{7}-\dfrac{23}{34}-\dfrac{4}{5}=\dfrac{3}{7}-\dfrac{23}{34}-\dfrac{4}{5}=-\dfrac{1247}{1190}\)

b:

Sửa đề: \(\dfrac{-5}{13}+\dfrac{4}{19}+\dfrac{-8}{13}+\dfrac{15}{19}+\dfrac{45}{6}\) 

\(=\dfrac{-5}{13}-\dfrac{8}{13}+\dfrac{4}{19}+\dfrac{15}{19}+\dfrac{45}{6}=\dfrac{9}{2}\)

\(a.\)

\(\dfrac{2}{9}+\dfrac{-3}{10}+-\dfrac{7}{10}=\dfrac{2}{9}-1=\dfrac{2}{9}-\dfrac{9}{9}=-\dfrac{7}{9}\)

\(b.\)

\(\dfrac{-11}{6}+\dfrac{2}{5}+\dfrac{-1}{6}=\left(-\dfrac{11}{6}+-\dfrac{1}{6}\right)+\dfrac{2}{5}=-2+\dfrac{2}{5}=\dfrac{-10}{5}+\dfrac{2}{5}=\dfrac{-8}{5}\)

\(c.\)

\(-\dfrac{5}{8}+\dfrac{12}{7}+\dfrac{13}{8}+\dfrac{2}{7}=\left(-\dfrac{5}{8}+\dfrac{13}{8}\right)+\left(\dfrac{12}{7}+\dfrac{2}{7}\right)=1+2=3\)

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

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

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

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

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

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: \(=\left(15-6-\dfrac{13}{18}\right):\dfrac{298}{27}-\dfrac{17}{8}:\dfrac{51}{40}\)

\(=\dfrac{149}{18}\cdot\dfrac{27}{298}-\dfrac{5}{3}=\dfrac{3}{2}-\dfrac{5}{3}=\dfrac{9-10}{6}=\dfrac{-1}{6}\)

b: \(=\dfrac{-16}{5}\cdot\dfrac{-15}{64}+\dfrac{-22}{15}:\dfrac{11}{2}\)

\(=\dfrac{3}{4}-\dfrac{4}{15}=\dfrac{29}{60}\)

c: \(=\dfrac{-7}{9}\left(\dfrac{4}{11}+\dfrac{7}{11}\right)+5+\dfrac{7}{9}=\dfrac{-7}{9}+\dfrac{7}{9}+5=5\)

d: \(=\dfrac{1}{2}\cdot\dfrac{4}{3}\cdot10\cdot\dfrac{1}{5}\cdot\dfrac{3}{4}=1\)

e: \(=\dfrac{4}{25}+\dfrac{11}{2}\cdot\dfrac{5}{2}+\dfrac{-23}{4}=\dfrac{204}{25}\)

`1)1/2:2/3 .... 2/3 : 1/2`

`=>1/2xx3/2 .... 2/3xx2`

`=>3/4 .... 4/3`

Vì `3/4 < 1` và `4/3>1` 

`=>3/4<4/3`

__

`4/7:2/5 ... 4/7 : 3/5`

`=>4/7xx5/2....4/7xx5/3`

`=>20/14...20/21`

`=>10/7...20/21`

Vì `10/7>1` và `20/21<1` 

`=>10/7>20/21`

__

`4/15:4/7....2/5xx10/3`

`=>4/15xx7/4...20/15`

`=>7/15...20/15`

Vì `7<20` nên `7/15<20/15`

__

`5/6...15/18-11/18`

`=>5/6...4/18`

Ta có : MSC : `18`

`5/6 = 15/18`

Vì `15>4` nên `5/6 > 4/18`

`2)2/3=(2xx6)/(3xx6)=12/18`

`7/9=(7xx7)/(9xx7)=49/63`

`6/5=(6xx3)/(5xx3)=18/15`

`2/3=(2xx5)/(3xx5)=10/15`

`5/9=(5xx5)/(9xx5)=25/45`

`49/56=(49:7)/(56:7)=7/8`

`6/8=(6xx7)/(8xx7)=42/56`

`2/9=(2xx7)/(9xx7)=14/63`

`49/56=(49:7)/(56:7)=7/8`

a: \(=\dfrac{3}{7}+\dfrac{7}{8}+\dfrac{4}{7}-\dfrac{15}{8}=1-1=0\)

b: \(=\dfrac{1}{2}+\dfrac{4}{5}+\dfrac{3}{8}-\dfrac{7}{40}+\dfrac{1}{2}\)

\(=1+\dfrac{32+15-7}{40}=1+1=2\)

a) \(\dfrac{5}{9}+\dfrac{13}{7}+\dfrac{15}{13}+\dfrac{8}{7}+\dfrac{4}{9}+\dfrac{11}{13}\)

\(=\left(\dfrac{5}{9}+\dfrac{4}{9}\right)+\left(\dfrac{13}{7}+\dfrac{8}{7}\right)+\left(\dfrac{15}{13}+\dfrac{11}{13}\right)\)

\(=1+3+2\)

\(=6\)

b) \(\dfrac{3}{5}+\dfrac{18}{9}+\dfrac{4}{10}\)

\(=0,6+2+0,4\)

\(=\left(0,6+0,4\right)+2\)

\(=1+2\)

\(=3\)

\(a\dfrac{5}{9}+\dfrac{4}{9}+\dfrac{13}{7}+\dfrac{8}{7}+\dfrac{15}{13}+\dfrac{11}{13}=\dfrac{9}{9}+\dfrac{21}{7}+\dfrac{26}{13}=1+3+2=6\)

\(b\dfrac{3}{5}+2+\dfrac{2}{5}=\dfrac{5}{5}+2=1+2=3\)