4: \(=\dfrac{2+3}{7}+\dfrac{1+6}{9}-\dfrac{5}{6}=\dfrac{5}{7}+\dfrac{7}{9}-\dfrac{5}{6}=\dfrac{83}{126}\)

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

6: \(=\dfrac{-3-28}{31}+\dfrac{-6-1}{17}+\dfrac{1-5}{25}=-1-\dfrac{7}{17}-\dfrac{4}{25}=-\dfrac{668}{425}\)

4. \(\dfrac{2}{7}+\dfrac{1}{9}+\dfrac{3}{7}+\dfrac{5}{9}+\dfrac{-5}{6}\)

\(=\left(\dfrac{2}{7}+\dfrac{3}{7}\right)+\left(\dfrac{1}{9}+\dfrac{5}{9}\right)+\dfrac{-5}{6}\)

\(=\dfrac{5}{7}+\dfrac{2}{3}+\dfrac{-5}{6}\)

\(=\dfrac{30+28+\left(-35\right)}{42}=\dfrac{23}{42}\)
 

5. \(\dfrac{-5}{7}+\dfrac{3}{4}+\dfrac{-1}{5}+\dfrac{-2}{7}+\dfrac{1}{4}\)

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

\(=\dfrac{-7}{7}+\dfrac{4}{4}+\dfrac{-1}{5}\)

\(=-1+1+\dfrac{-1}{5}\)

\(=-\dfrac{1}{5}\)
 

6. \(\dfrac{-3}{31}+\dfrac{-6}{17}+\dfrac{1}{25}+\dfrac{-28}{31}+\dfrac{-1}{17}+\dfrac{-1}{5}\)

\(=\left(\dfrac{-3}{31}+\dfrac{-28}{31}\right)+\left(\dfrac{-6}{17}+\dfrac{-1}{17}\right)+\dfrac{1}{25}+\dfrac{-1}{5}\)

\(=\dfrac{-31}{31}+\dfrac{-7}{17}+\dfrac{1}{25}+\dfrac{-1}{5}\)

\(=-1+\dfrac{-7}{17}+\dfrac{1}{25}+\dfrac{-1}{5}\)

\(=\dfrac{-425+\left(-175\right)+17+\left(-85\right)}{425}=\dfrac{-668}{425}\)

chx làm :)))

à nhìn nhầm

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

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

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

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

A=4/63

`@` `\text {Ans}`

`\downarrow`

`a.`

`A=(1/2-7/13-1/3)+(-6/13+1/2+1 1/3)`

`= 1/2 - 7/13 - 1/3 - 6/13 + 1/2 + 1 1/3`

`= (1/2 + 1/2) + (-7/13 - 6/13) + (-1/3 + 1 1/3) `

`= 1 - 1 + 1`

`= 1`

`b.`

`B=0,75+2/5+(1/9-1 1/2+5/4)`

`= 3/4 + 2/5 + 1/9 - 3/2 + 5/4`

`= (3/4+5/4)+ 1/9 + 2/5 - 3/2`

`= 2 + 1/9 - 11/10`

`= 19/9 - 11/10`

`= 91/90`

`c.`

`(-5/9).3/11+(-13/18).3/11`

`= 3/11*[(-5/9) + (-13/18)]`

`= 3/11*(-23/18)`

`= -23/66`

`d.`

`(-2/3).3/11+(-16/9).3/11`

`= 3/11* [(-2/3) + (-16/9)]`

`= 3/11*(-22/9)`

`= -2/3`

`e.`

`(-1/4).(-2/13)-7/24.(-2/13)`

`= (-2/13)*(-1/4-7/24)`

`= (-2/13)*(-13/24)`

`= 1/12`

`f.`

`(-1/27).3/7+(5/9).(-3/7)`

`= 3/7*(-1/27 - 5/9)`

`= 3/7*(-16/27)`

`= -16/63`

`g.`

`(-1/5+3/7):2/11+(-4/5+4/7):2/11`

`=[(-1/5+3/7)+(-4/5+4/7)] \div 2/11`

`= (-1/5+3/7 - 4/5 + 4/7) \div 2/11`

`= [(-1/5-4/5)+(3/7+4/7)] \div 2/11`

`= (-1+1) \div 2/11`

`= 0 \div 2/11 = 0`

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

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

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

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