Bài 1: 

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

\(B=\dfrac{2}{15}+\dfrac{5}{9}-\dfrac{6}{9}=\dfrac{2}{15}-\dfrac{1}{9}=\dfrac{18-15}{135}=\dfrac{3}{135}=\dfrac{1}{45}\)

\(C=\dfrac{-1}{5}+\dfrac{1}{4}-\dfrac{3}{4}=\dfrac{-1}{5}-\dfrac{1}{2}=\dfrac{-7}{10}\)

Bài 2: 

a: \(=\dfrac{1}{5}+\dfrac{1}{2}+\dfrac{2}{5}-\dfrac{3}{5}+\dfrac{2}{21}-\dfrac{10}{21}+\dfrac{3}{20}\)

\(=\left(\dfrac{1}{5}+\dfrac{2}{5}-\dfrac{3}{5}\right)+\left(\dfrac{2}{21}-\dfrac{10}{21}\right)+\left(\dfrac{1}{2}+\dfrac{3}{20}\right)\)

\(=\dfrac{-8}{21}+\dfrac{13}{20}=\dfrac{113}{420}\)

b: \(B=\dfrac{21}{23}-\dfrac{21}{23}+\dfrac{125}{93}-\dfrac{125}{143}=\dfrac{6250}{13299}\)

Bài 3:

\(\dfrac{7}{3}-\dfrac{1}{2}-\left(-\dfrac{3}{70}\right)=\dfrac{7}{3}-\dfrac{1}{2}+\dfrac{3}{70}=\dfrac{490}{210}-\dfrac{105}{210}+\dfrac{9}{210}=\dfrac{394}{210}=\dfrac{197}{105}\)

\(\dfrac{5}{12}-\dfrac{3}{-16}+\dfrac{3}{4}=\dfrac{5}{12}+\dfrac{3}{16}+\dfrac{3}{4}=\dfrac{20}{48}+\dfrac{9}{48}+\dfrac{36}{48}=\dfrac{65}{48}\)

Bài 4:

 \(\dfrac{3}{4}-x=1\)

\(\Rightarrow-x=1-\dfrac{3}{4}\)

\(\Rightarrow x=-\dfrac{1}{4}\)

Vậy: \(x=-\dfrac{1}{4}\)

\(x+4=\dfrac{1}{5}\)

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

\(\Rightarrow x=-\dfrac{19}{5}\)

Vậy: \(x=-\dfrac{19}{5}\)

\(x-\dfrac{1}{5}=2\)

\(\Rightarrow x=2+\dfrac{1}{5}\)

\(\Rightarrow x=\dfrac{11}{5}\)

Vậy: \(x=\dfrac{11}{5}\)

\(x+\dfrac{5}{3}=\dfrac{1}{81}\)

\(\Rightarrow x=\dfrac{1}{81}-\dfrac{5}{3}\)

\(\Rightarrow x=-\dfrac{134}{81}\)

Vậy: \(x=-\dfrac{134}{81}\)

a, \(-1,8:\left(1-\frac{3}{4}\right)\)

\(=-\frac{9}{5}:\frac{1}{4}\)

\(=-\frac{36}{5}\)

b, \(\left(\frac{7}{8}-\frac{3}{4}\right)\cdot1\frac{1}{3}-\frac{2}{7}\cdot\left(3,5\right)^2\)

\(=\left(\frac{7}{8}-\frac{6}{8}\right)\cdot\frac{4}{3}-\frac{2}{7}\cdot12.25\)

\(=\frac{1}{8}\cdot\frac{4}{3}-\frac{2}{7}\cdot\frac{49}{4}\)

\(=\frac{1}{6}-\frac{7}{2}\)

\(=\frac{1}{6}-\frac{21}{6}\)

\(=-\frac{20}{6}\)

\(=-\frac{10}{3}\)

lần này mới được à

? Lần này mới được

a: =6x2/5-3/10

=12/5-3/10

=24/10-3/10=21/10

b: =1/2+4/3=3/6+8/6=11/6

c: =5/6+1/4=10/12+3/12=13/12

`a, 3/4 + 1/2 xx 7/2`

`= 3/4 + 7/4`

`=10/4`

`=5/2`

`b, 6/15 - 1/3 : 5/3`

`= 6/15 - 1/3 xx 3/5`

`= 6/15 - 3/15`

`= 3/15`

`=1/5`

`c, x-4/9 = 3/7 : 9/4`

`=> x-4/9= 3/7 xx 4/9`

`=> x-4/9= 12/63`

`=> x-4/9=4/21`

`=> x= 4/21 +4/9`

`=>x= 40/63`

`d, 7/9 xx 3/5 -1/2=1/5`

`->` sao lại bằng có `x` ko vậy ạ?

`a,`

`3/4+1/2 \times 7/2=3/4+7/4=10/4=5/2`

`b,`

`6/15 - 1/3 \div 5/3=6/15-1/5=1/5`

`c,` Tìm x?

`x-4/9=3/7 \div 9/4`

`x-4/9=4/21`

`x=4/21+4/9`

`x=40/63`

`d, 7/9x \times 3/5-1/2=1/5`

`7/9x \times 3/5=1/5+1/2`

`7/9x \times 3/5=7/10`

`7/9x=7/10 \div 3/5`

`7/9x=7/6`

`x=7/6 \div 7/9=3/2`

A = ( 6 : 3/5 - 7/6  * 6/7 ) : ( 21/5 * 10/11 + 57/11 )

A = ( 10 -  1 )  : ( 42/11 + 57/11)

A =    9   :  9

A =       1

B = 59 /10 : 3/2 - ( 7/3 * 9/2 - 2 * 7/3 ) : 7/4

B =   59/15  - (  21/2 -  14/3 )  : 7/4

B =    59/15 - 35/6 : 7/4

B  =    59/15 - 10/3

B  =         3/5

cảm ơn bạn nhiều nha

còn câu C thì sao

2: 

a: =>x=1/2+1/6=3/6+1/6=4/6=2/3

b: =>x+3,5=7,8-6,3=1,5

=>x=1,5-3,5=-2

c: =>(35-x)/-105=1/5

=>35-x=-21

=>x=56

1

c hình như tính bình thường thôi:v

d

= 1,75 : 5 + 2,5 . (16 – 4 . 4,1)

= 1,75 : 5 + 2,5 . (16 – 16,4)

= 1,75 : 5 + 2,5 . (−0,4)

= 0,35 − 1

= −0,65.

2

a

\(\Rightarrow x=\dfrac{1}{2}+\dfrac{1}{6}=\dfrac{3}{6}+\dfrac{1}{6}=\dfrac{4}{6}=\dfrac{2}{3}\)

b

\(\Rightarrow x+\dfrac{35}{10}=\dfrac{78}{10}-\dfrac{63}{5}:2\\ \Rightarrow x+\dfrac{35}{10}=\dfrac{78}{10}-\dfrac{63}{10}=\dfrac{15}{10}\\ \Rightarrow x=\dfrac{15}{10}-\dfrac{35}{10}=-2\)

c không rõ đề:v

\(1,\dfrac{2}{-11}.6\dfrac{2}{7}+\dfrac{3}{7}:3+5\\ =\dfrac{2}{-11}.\dfrac{44}{7}+\dfrac{3}{7}\times\dfrac{1}{3}+5\\ =-\dfrac{88}{77}+\dfrac{3}{21}+5\\ =-\dfrac{8}{7}+\dfrac{1}{7}+5\\ =-1+5\\ =4\)

\(d,1,75:5+2,5.\left(4^2-4.4,1\right)\\ =0,35+2,5.\left(16-16,4\right)\\ =0,35+2,5.0,4\\ =0,35+1\\ =1,35\)

`2,`

`a, 1/2 -x=-1/6`

`=> x= 1/2 -(-1/6)`

`=> x= 1/2 +1/6`

`=> x= 3/6+1/6`

`=>x= 4/6`

`=>x=2/3`

Vậy `x=2/3`

`b, x+3,5 =7,8 -12,6 :2`

`=> x+3,5 = 7,8 - 6,3`

`=> x+3,5=1,5`

`=> x= 1,5-3,5`

`=>x=-2`

Vậy `x=-2`

\(c,\dfrac{7}{35}=\dfrac{35-x}{-105}\\ \Rightarrow7.\left(-105\right)=35.\left(35-x\right)\\ \Rightarrow-735=1225-35x\\ \Rightarrow-735-1225=-35x\\ \Rightarrow-1960=-35x\\ \Rightarrow x=-1960:\left(-35\right)\\ \Rightarrow x=56\)

Vậy `x=56`