\(\dfrac{9}{5}-\dfrac{3}{4}:\dfrac{5}{6}=\dfrac{9}{5}-\dfrac{9}{10}=\dfrac{9}{10}\)

\(x=\dfrac{7}{9}:\dfrac{5}{7}=\dfrac{49}{45}\)

\(=\dfrac{9}{5}-\dfrac{3}{4}x\dfrac{6}{5}=\dfrac{9}{5}-\dfrac{9}{10}=\dfrac{18}{10}-\dfrac{9}{10}=\dfrac{9}{10}\)

\(x=\dfrac{7}{9}:\dfrac{5}{7}\)

\(x=\dfrac{49}{45}\)

9/5-3/4:5/6=9/5-9/10=9/10

7/9:x=5/7

x=7/9:5/7

x=49/45

`(6/11 +5/11) xx 3/7`

`= 11/11xx 3/7`

`=1xx3/7`

`=3/7`

__

`3/5 xx 7/9 - 3/5 xx 2/9`

`= 3/5 xx (7/9-2/9)`

`= 3/5 xx 5/9`

`= 15/45`

`= 1/3`

__

`(6/7 -4/7):2/5`

`= 2/7 : 2/5`

`= 2/7 xx 5/2`

`= 10/14`

`= 5/7`

\(\left(\dfrac{6}{11}+\dfrac{5}{11}\right)\times\dfrac{3}{7}\)

\(=1\times\dfrac{3}{7}=\dfrac{3}{7}\)

___________

\(\dfrac{3}{5}\times\dfrac{7}{9}-\dfrac{3}{5}\times\dfrac{2}{9}\)

\(=\dfrac{3}{5}\times\left(\dfrac{7}{9}-\dfrac{2}{9}\right)\)

\(=\dfrac{3}{5}\times\dfrac{5}{9}\)

\(=\dfrac{1}{3}\)

__________

\(\left(\dfrac{6}{7}-\dfrac{4}{7}\right):\dfrac{2}{5}\)

\(=\dfrac{2}{7}\times\dfrac{5}{2}\)

\(=\dfrac{5}{7}\)

\(#WendyDang\)

\(\left(\dfrac{6}{11}+\dfrac{5}{11}\right)\times\dfrac{3}{7}\)

\(=\dfrac{11}{11}\times\dfrac{3}{7}\\ =1\times\dfrac{3}{7}=\dfrac{3}{7}\)

_____

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

_________

\(\left(\dfrac{6}{7}-\dfrac{4}{7}\right):\dfrac{2}{5}\\ =\dfrac{2}{7}\times\dfrac{5}{2}\\ =\dfrac{10}{14}\\ =\dfrac{5}{7}\)

\(=\dfrac{3}{4}-\dfrac{5}{6}\times\dfrac{7}{24}\times\dfrac{12}{7}=\dfrac{3}{4}-\dfrac{5}{12}=\dfrac{1}{3}\)

\(\dfrac{3}{4}-\dfrac{5}{6}\left(\dfrac{1}{6}+\dfrac{1}{8}\right):\dfrac{7}{12}\)

\(=\dfrac{3}{4}-\dfrac{5}{6}\cdot\dfrac{7}{24}\cdot\dfrac{12}{7}\)

\(=\dfrac{3}{4}-\dfrac{5}{12}\)

\(=\dfrac{4}{12}=\dfrac{1}{3}\)

1/3

\(\dfrac{5}{8}:\dfrac{3}{4}-\dfrac{1}{6}=\dfrac{5}{8}\times\dfrac{4}{3}-\dfrac{1}{6}=\dfrac{5}{6}-\dfrac{1}{6}=\dfrac{4}{6}=\dfrac{2}{3}\)

\(\left(\dfrac{3}{14}+\dfrac{1}{2}\right)\times\dfrac{7}{5}=\left(\dfrac{3}{14}+\dfrac{7}{14}\right)\times\dfrac{7}{15}=\dfrac{10}{14}\times\dfrac{7}{15}=\dfrac{5}{7}\times\dfrac{7}{15}=\dfrac{5}{12}=\dfrac{1}{3}\)

`1/3-2/15+14/15`

`=5/15-2/15+14/15`

`=17/15`

`3/5+4-6/7`

`=21/35+140/35-30/35`

`=131/35`

giúp mình với ạ, thanks!

`1/3- 2/15 + 14/15`

`= 5/15 -2/15 + 14/15`

`=(5-2+14)/15`

`= 17/15`

`--`

`3/5 + 4 -6/7`

`= 3/5 + 20/5 - 6/7`

`= 23/5 - 6/7`

`= 131/35`

a) ( 42- 6) : 1/2 + 7/6

= 36 : 1/2 + 7/6

= 72 + 7/6

= 439/6

b) 18/5 - 2/7 + 1/3

= 383/105

a) = ( 42 - 6) : 1/2 + 7/6 

= 36 : 1/2 + 7/6

= 72 + 7/6

= 439/6

b) = 18/5 - 2/7 + 1/3

= 116/35 + 1/3

= 383/105

\(a.=\left(42-6\right):\dfrac{1}{2}+\dfrac{7}{6}\)

\(=36:\dfrac{1}{2}+\dfrac{7}{6}\)

\(=\)\(72+\dfrac{7}{6}\)

\(=\dfrac{439}{6}\)

GIÚP ĐI XIN ĐÓ AI ĐÓA TRẢ LỜI DÙM CẦN GẤP

x = 28/5-5

x= 3/5

1/3 :3/4 = 1/3 x 4/3= 4/9

Bài 1: ĐKXĐ:`x + 3 ne 0` và `x^2+ x-6 ne 0 ; 2-x ne 0`

`<=> x ne -3 ; (x-2)(x+3) ne 0 ; x ne2`

`<=>x ne -3 ; x ne 2`

b) Với `x ne - 3 ; x ne 2` ta có:

`P= (x+2)/(x+3)  - 5/(x^2 +x -6) + 1/(2-x)`

`P = (x+2)/(x+3) - 5/[(x-2)(x+3)] + 1/(2-x)`

`= [(x+2)(x-2)]/[(x-2)(x+3)] - 5/[(x-2)(x+3)] - (x+3)/[(x-2)(x+3)]`

`= (x^2 -4)/[(x-2)(x+3)] - 5/[(x-2)(x+3)] - (x+3)/[(x-2)(x+3)]`

`=(x^2 - 4 - 5 - x-3)/[(x-2)(x+3)]`

`= (x^2 - x-12)/[(x-2)(x+3)]`

`= [(x-4)(x+3)]/[(x-2)(x+3)]`

`= (x-4)/(x-2)`

Vậy `P= (x-4)/(x-2)` với `x ne -3 ; x ne 2`

c) Để `P = -3/4`

`=> (x-4)/(x-2) = -3/4`

`=> 4(x-4) = -3(x-2)`

`<=>4x -16 = -3x + 6`

`<=> 4x + 3x = 6 + 16`

`<=> 7x = 22`

`<=> x= 22/7` (thỏa mãn ĐKXĐ)

Vậy `x = 22/7` thì `P = -3/4`

d) Ta có: `P= (x-4)/(x-2)`

`P= (x-2-2)/(x-2)`

`P= 1 - 2/(x-2)`

Để P nguyên thì `2/(x-2)` nguyên

`=> 2 vdots x-2`

`=> x -2 in Ư(2) ={ 1 ;2 ;-1;-2}`

+) Với `x -2 =1 => x= 3` (thỏa mãn ĐKXĐ)

+) Với `x -2 =2 => x= 4`  (thỏa mãn ĐKXĐ)

+) Với `x -2 = -1=> x= 1` (thỏa mãn ĐKXĐ)

+) Với `x -2 = -2 => x= 0`(thỏa mãn ĐKXĐ)

Vậy `x in{ 3 ;4; 1; 0}` thì `P` nguyên

e) Từ `x^2 -9 =0`

`<=> (x-3)(x+3)=0`

`<=> x= 3` hoặc `x= -3`

+) Với `x=3` (thỏa mãn ĐKXĐ) thì:

`P  = (3-4)/(3-2)`

`P= -1/1`

`P=-1`

+) Với `x= -3` thì không thỏa mãn ĐKXĐ

Vậy với x= 3 thì `P= -1`

\(=\dfrac{3\left(\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{7}+\dfrac{1}{11}\right)}{13\left(\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{17}+\dfrac{1}{11}\right)}=\dfrac{3}{13}\)

\(\dfrac{\dfrac{3}{4}-\dfrac{3}{5}+\dfrac{3}{7}+\dfrac{3}{11}}{\dfrac{13}{4}-\dfrac{13}{5}+\dfrac{13}{7}+\dfrac{13}{11}}\)

\(=\dfrac{3\left(\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{7}+\dfrac{1}{11}\right)}{13\left(\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{7}+\dfrac{1}{11}\right)}\)

\(=\dfrac{3}{13}\)