c) Ta có: \(\dfrac{3}{5}+\dfrac{-5}{20}+\dfrac{30}{75}+\dfrac{-7}{4}\)

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

\(=1-2=-1\)

Giải:

a)-1/12+4/3=-1/12+16/12=15/12=5/4

b)(-4/14-3/15)-(1/5-20/35-(-1)).7

=-17/35-22/35.7

=-17/35-22/5

=-171/35

c)3/5+-5/20+30/75+-7/4

=3/5+-1/4+2/5+-7/4

=(3/5+2/5)+(-1/4+-7/4)

=1+-2

=-1

d)5/6.-12/14+7/13

=-5/7+7/13

=-16/91

e)2/-9-5/-36-1/4

=-1/12-1/4

=-1/3

f)2/23+-5/12+7/18+21/23+-7/12

=(2/23+21/23)+(-5/12+-7/12)+7/18

=1+-1+7/18

=7/18

a)137/140,b)1/7 

nhớ k mình nha!

a)137/140

b)1/4

Lời giải:
a.

$-18: \frac{3}{5}=-18.\frac{5}{3}=-30$

b.

$\frac{3}{4}:(-9)=\frac{3}{4}.\frac{-1}{9}=\frac{-1}{12}$

c.

$\frac{13}{20}-\frac{6}{7}: \frac{10}{21}=\frac{13}{20}-\frac{6}{7}.\frac{21}{10}$

$=\frac{13}{20}-\frac{9}{5}=\frac{13}{20}-\frac{36}{20}=\frac{-23}{20}$

d.

$\frac{-21}{5}: (\frac{7}{3}.\frac{7}{5})=\frac{-21}{5}: \frac{49}{15}$

$=\frac{-21}{5}.\frac{15}{49}=\frac{-9}{7}$

e.

$(\frac{-2}{5}+\frac{1}{4}): (1-\frac{2}{5})$

$=\frac{-3}{20}: \frac{3}{5}=\frac{-1}{4}$

\(1,-\dfrac{4}{7}+\dfrac{2}{3}\times\dfrac{-9}{14}\)

\(=\dfrac{-4}{7}+\dfrac{-18}{42}\)

\(=\dfrac{-4\times6}{7\times6}+\dfrac{-18}{42}\)

\(=\dfrac{-20}{42}+\dfrac{-18}{42}\)

\(=-\dfrac{38}{42}\)

\(=-\dfrac{19}{21}\)

\(2,\dfrac{17}{13}-\left(\dfrac{4}{13}-11\right)\)

\(=\dfrac{17}{13}-\dfrac{4}{13}+11\)

\(=\dfrac{13}{13}+11\)

\(=1+11\)

\(=12\)

\(3,8\dfrac{2}{7}-\left(3\dfrac{4}{9}+4\dfrac{2}{7}\right)\)

\(=\dfrac{58}{7}-\left(\dfrac{31}{9}+\dfrac{30}{7}\right)\)

\(=\dfrac{58}{7}-\dfrac{31}{9}-\dfrac{30}{7}\)

\(=\dfrac{58}{7}-\dfrac{30}{7}-\dfrac{31}{9}\)

\(=\dfrac{28}{7}-\dfrac{31}{9}\)

\(=\dfrac{28\times9}{7\times9}-\dfrac{31\times7}{9\times7}\)

\(=\dfrac{252}{63}-\dfrac{217}{63}\)

\(=\dfrac{35}{63}\)

\(=\dfrac{5}{9}\)

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

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

\(=\left(\dfrac{2\times2}{3\times2}-\dfrac{3\times3}{2\times3}\right):\dfrac{4}{3}+\dfrac{1}{2}\)

\(=\left(\dfrac{4}{6}-\dfrac{9}{6}\right):\dfrac{4}{3}+\dfrac{1}{2}\)

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

\(=\dfrac{-5}{6}\times\dfrac{3}{4}+\dfrac{1}{2}\)

\(=\dfrac{-15}{24}+\dfrac{1}{2}\)

\(=\dfrac{-15}{24}+\dfrac{1\times12}{2\times12}\)

\(=\dfrac{-15}{24}+\dfrac{12}{24}\)

\(=\dfrac{-3}{24}\)

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

\(6,\dfrac{-5}{13}+\dfrac{2}{5}+\dfrac{-8}{13}+\dfrac{3}{5}-\dfrac{3}{7}\)

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

\(=\dfrac{-13}{13}+\dfrac{5}{5}-\dfrac{3}{7}\)

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

\(=-\dfrac{3}{7}\)

\(7,\dfrac{6}{5}\times\dfrac{3}{7}+\dfrac{6}{5}:\dfrac{7}{10}+\dfrac{6}{5}\)

\(=\dfrac{6}{5}\times\dfrac{3}{7}+\dfrac{6}{5}\times\dfrac{10}{7}+\dfrac{6}{5}\)

\(=\dfrac{6}{5}\times\left(\dfrac{3}{7}+\dfrac{10}{7}+1\right)\)

\(=\dfrac{6}{5}\times\left(\dfrac{3}{7}+\dfrac{10}{7}+\dfrac{1\times7}{1\times7}\right)\)

\(=\dfrac{6}{5}\times\left(\dfrac{3}{7}+\dfrac{10}{7}+\dfrac{7}{7}\right)\)

\(=\dfrac{6}{5}\times\dfrac{20}{7}\)

\(=\dfrac{120}{35}\)

\(=\dfrac{24}{7}\)

`@` `\text {Ans}`

`\downarrow`

`a.`

\(0,3-\dfrac{4}{9}\div\dfrac{4}{3}\cdot\dfrac{6}{5}+1\)

`=`\(0,3-\dfrac{1}{3}\cdot\dfrac{6}{5}+1\)

`=`\(0,3-0,4+1\)

`= -0,1 + 1`

`= 0,9`

`b.`

\(1+2\div\left(\dfrac{2}{3}-\dfrac{1}{6}\right)\cdot\left(-2,25\right)\)

`=`\(1+2\div\dfrac{1}{2}\cdot\left(-2,25\right)\)

`=`\(1+4\cdot\left(-2,25\right)\)

`= 1+ (-9) = -8`

`c.`

\(\left[\left(\dfrac{1}{4}-0,5\right)\cdot2+\dfrac{8}{3}\right]\div2\)

`=`\(\left(-\dfrac{1}{4}\cdot2+\dfrac{8}{3}\right)\div2\)

`=`\(\left(-\dfrac{1}{2}+\dfrac{8}{3}\right)\div2\)

`=`\(\dfrac{13}{6}\div2\)

`=`\(\dfrac{13}{12}\)

`d.`

\(\left[\left(\dfrac{3}{8}-\dfrac{5}{12}\right)\cdot6+\dfrac{1}{3}\right]\cdot4\)

`=`\(\left(-\dfrac{1}{24}\cdot6+\dfrac{1}{3}\right)\cdot4\)

`=`\(\left(-\dfrac{1}{4}+\dfrac{1}{3}\right)\cdot4\)

`=`\(\dfrac{1}{12}\cdot4=\dfrac{1}{3}\)

`e.`

\(\left(\dfrac{4}{5}-1\right)\div\dfrac{3}{5}-\dfrac{2}{3}\cdot0,5\)

`=`\(-\dfrac{1}{5}\div\dfrac{3}{5}-\dfrac{1}{3}\)

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

`f.`

\(0,8\div\left\{0,2-7\left[\dfrac{1}{6}+\left(\dfrac{5}{21}-\dfrac{5}{14}\right)\right]\right\}\)

`=`\(0,8\div\left[0,2-7\left(\dfrac{1}{6}-\dfrac{5}{42}\right)\right]\)

`=`\(0,8\div\left(0,2-7\cdot\dfrac{1}{21}\right)\)

`=`\(0,8\div\left(0,2-\dfrac{1}{3}\right)\)

`= 0,8 \div (-2/15)`

`=-6`

`@` `yHGiangg.`

Mk ko ghi lại đề nhé

1 = \(350-4.19+4.7\)

\(=350-4.\left(19+7\right)\)

\(=350+4.26\)

\(=350-104=246\)

2 Câu này mình vẫn chưa hiểu là bạn ghi 27  3/5 tức là 27.3/5 hay 27\(\dfrac{3}{5}\)nên mk bỏ qua nhé

3 Cái đoạn 1 1/3 y chang câu 2 

4 \(=\dfrac{2}{3}+\dfrac{1}{3}.\dfrac{7}{18}.\dfrac{12}{7}\)

\(=\dfrac{2}{3}+\dfrac{7}{54}.\dfrac{12}{7}\)

\(=\dfrac{2}{3}+\dfrac{2}{9}=\dfrac{8}{9}\)

5 \(=2^2+10^2-\left[9.\left(112:8\right)-11\right].1\)

\(=4+100-\left(9.14-11\right)\)

\(=104-115\)\(=-11\)

6 Đoạn 6 3/5 y chang 2,3

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

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

\(=\dfrac{-196}{588}+\dfrac{-84}{588}-\dfrac{-21}{588}\)\(=\dfrac{259}{588}\)

Làm tạm tạm thôi xl vì có 3 câu ko hiểu 

tách ra bn

khó nhìn quá:V

Chóng mặttttttttttttttt!

ko xem dc de ban oi :/

2