a: \(A=\dfrac{3^6\cdot3^8\cdot5^4-3^{13}\cdot5^{13}\cdot5^{-9}}{3^{12}\cdot5^6+5^6\cdot3^{12}}\)

\(=\dfrac{3^{14}\cdot5^4-3^{13}\cdot5^4}{2\cdot3^{12}\cdot5^6}\)

\(=\dfrac{3^{13}\cdot5^4\cdot\left(3-1\right)}{2\cdot3^{12}\cdot5^6}=\dfrac{3}{5^2}=\dfrac{3}{25}\)

c: \(C=\dfrac{\dfrac{27}{64}+\dfrac{125}{64}-5\cdot\dfrac{16-15}{12}}{\dfrac{25}{64}+\dfrac{4}{9}-\dfrac{5}{6}}\)

\(=\dfrac{47}{24}:\dfrac{1}{576}=47\cdot24=1128\)

a) Ta có: \(a\left(-\dfrac{3}{2}\right)+a\cdot\dfrac{1}{4}-a\cdot\dfrac{5}{6}\)

\(=a\left(-\dfrac{3}{2}+\dfrac{1}{4}-\dfrac{5}{6}\right)\)

\(=a\left(\dfrac{-18}{12}+\dfrac{3}{12}-\dfrac{10}{12}\right)\)

\(=a\cdot\dfrac{-25}{12}\)(1)

Thay \(a=\dfrac{3}{5}\) vào biểu thức (1), ta được:

\(\dfrac{3}{5}\cdot\dfrac{-25}{12}=\dfrac{-75}{60}=\dfrac{-5}{4}\)

35⋅−2512=−7560=−54

` a/`

` 2 - 1 5/6 + 2 2/3 = 2 - 11/6 - 8/3 = 1/6+ 8/3 = 1/6 + 16/6 = 17/6 `

`b/`

`5/9 xx ( 2 5/6 - 1 2/3 ) = 5/9 xx ( 17/6 - 5/3 ) = 5/9 xx 7/6 = 35/54 `

`c/`

` 1 1/3 : ( 2 + 1 1/6 : 2 5/6 ) `

`= 4/3 :  ( 2 + 7/6 : 17/6 ) `

`= 4/3 : ( 2 + 7/6 xx 6/17 )`

`= 4/3 : ( 2 + 7/17 ) `

`= 4/3 : ( 34/17 + 7/17 ) `

`= 4/3 : 41/17 `

`= 4/3 xx 17/41 `

`= 68/123`

` d/` 

` 2 3/5 : 3/4 xx 1 4/5 = 13/5 xx 4/3 xx 9/5 =52/15 xx 9/5 = 156/25`

a) `1/3 - 1/4 : 2/5 = 1/3 - 5/8 = -7/24`

b) `6/7-(5/6+1/3)-(2/3+1/7) = 6/7-5/6-1/3-2/3-1/7`

`=(6/7-1/7)-(1/3+2/3)-5/6`

`=5/7-1-5/6`

`=-47/42`

c) `-5/9 . 2/5 + 4 5/9 + 5/9 . (-3/5)`

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

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

`=-5/9 . 0 +4`

`=4`

d) 3 1/2 - (5 4/7 - 1 1/2) : 0,75`

`=7/2 - (39/7 - 3/2) : 3/4`

`= 7/2 - 57/14 : 3/4`

`=7/2 - 38/7`

`=-27/14`

De tke nma khonq bt lamm

Nguu

ét ô ét

a.25/27                                                                                                                 b.0                          c.0

\(a,0,5+\dfrac{1}{3}+0,4+\dfrac{5}{7}-\dfrac{1}{6}-\dfrac{4}{35}\\ =\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{2}{5}+\dfrac{5}{7}-\dfrac{1}{6}-\dfrac{4}{35}\\ =\dfrac{5}{6}+\dfrac{39}{35}-\dfrac{1}{6}-\dfrac{4}{35}\\ =\left(\dfrac{5}{6}-\dfrac{1}{6}\right)+\left(\dfrac{39}{35}-\dfrac{4}{35}\right)\\ =\dfrac{2}{3}+1\\ =\dfrac{4}{3}.\)

\(b,\left(3-\dfrac{1}{4}+\dfrac{2}{3}\right)-\left(5+\dfrac{1}{3}-\dfrac{6}{5}\right)-\left(-6-\dfrac{7}{4}+\dfrac{3}{2}\right)\\ =3-\dfrac{1}{4}+\dfrac{2}{3}-5-\dfrac{1}{3}+\dfrac{6}{5}+6+\dfrac{7}{4}-\dfrac{3}{2}\\ =\left(3-5+6\right)+\left(-\dfrac{1}{4}+\dfrac{7}{4}\right)+\left(\dfrac{2}{3}-\dfrac{1}{3}\right)+\left(\dfrac{6}{5}+\dfrac{7}{4}\right)\\ =4-\dfrac{3}{2}+\dfrac{1}{3}+\dfrac{59}{20}\\ =\dfrac{5}{2}+\dfrac{1}{3}+\dfrac{59}{20}\\ =\dfrac{17}{6}+\dfrac{59}{20}\\ =\dfrac{347}{60}.\)

\(c,\dfrac{1}{3}-\dfrac{3}{4}-\left(-\dfrac{3}{5}\right)+\dfrac{1}{64}-\dfrac{2}{9}-\dfrac{1}{36}+\dfrac{1}{15}\\ =\dfrac{1}{3}+\dfrac{3}{4}+\dfrac{3}{5}+\dfrac{1}{64}-\dfrac{2}{9}-\dfrac{1}{36}+\dfrac{1}{15}\\ =\left(\dfrac{1}{3}-\dfrac{2}{9}\right)+\left(\dfrac{3}{4}-\dfrac{1}{36}\right)+\left(\dfrac{3}{5}+\dfrac{1}{15}\right)+\dfrac{1}{64}\\ =\dfrac{1}{9}+\dfrac{13}{18}+\dfrac{2}{3}+\dfrac{1}{64}\\ =\dfrac{3}{2}+\dfrac{1}{64}\\ =\dfrac{65}{64}.\)

Lỗi sai ở câu (a) và câu (d). Sửa lại như sau

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

d) \(-\dfrac{2}{3}+\dfrac{2}{-5}=-\dfrac{2}{3}+-\dfrac{2}{5}=-\dfrac{10}{15}+-\dfrac{6}{15}=-\dfrac{16}{15}\)

a) x=4/2

b) x=1/12

c) 24/35

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

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

\(\Rightarrow x=2\)

b,\(\dfrac{3}{4}-x=\dfrac{2}{3}\)

\(x=\dfrac{3}{4}-\dfrac{2}{3}\)

\(\Rightarrow x=\dfrac{1}{12}\)

c,\(x:\dfrac{4}{5}=\dfrac{6}{7}\)

\(x=\dfrac{6}{7}.\dfrac{4}{5}\)

\(\Rightarrow\dfrac{24}{35}\)

a)1/2+x=5/2                                          b)3/4-x=2/3

    x=5/2-1/2=4/2=2                                  x=3/4-2/3=1/12

c)x:4/5=6/7

   x=6/7.4/5=24/35

a) Đ

b) S 

\(\dfrac{7}{10}-\dfrac{1}{5} \\ =\dfrac{7}{10}-\dfrac{2}{10}\\ =\dfrac{5}{10}=\dfrac{1}{2}\)

c) S

\(\dfrac{5}{4}+\dfrac{5}{12}\\ =\dfrac{15}{12}+\dfrac{5}{12}\\ =\dfrac{20}{12}=\dfrac{5}{3}\)

d) Đ