https://khoahoc.vietjack.com/question/661489/bieu-thuc-a-3-45-9-6-7-4-3-12-5-11-3-co-gia-tri-3-11

`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`

Bài 1: 

a, \(\dfrac{-x-2}{3}\) = - \(\dfrac{6}{7}\)

      - \(x\) - 2 = - \(\dfrac{18}{7}\)

         \(x\)      = - 2 + \(\dfrac{18}{7}\)

         \(x\)      = - \(\dfrac{4}{7}\)

Bài b,  \(\dfrac{4}{7-x}\) = \(\dfrac{1}{3}\)

            12 = 7 - \(x\)

            \(x\)  = 7 - 12 

            \(x\)  = -5 

Lời giải:

a.

$\frac{5}{15}-\frac{1}{6}\times \frac{2}{5}=\frac{5}{15}-\frac{1}{15}=\frac{4}{15}$

b.

$\frac{8}{24}+\frac{3}{4}:\frac{1}{8}=\frac{1}{3}+6=\frac{19}{3}$

c.

$\frac{1}{7}: \frac{2}{8}-\frac{1}{7}=\frac{1}{7}\times 4-\frac{1}{7}$

$=\frac{1}{7}\times (4-1)=\frac{1}{7}\times 3=\frac{3}{7}$

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 = ( 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

giúp mình mọi người ơi

a: \(A=0x^2y^4z+\dfrac{7}{2}x^2y^4z-\dfrac{2}{5}x^2y^4z=\dfrac{31}{10}x^2y^4z=\dfrac{31}{10}\cdot2^2\cdot\dfrac{1}{16}\cdot\left(-1\right)=-\dfrac{31}{40}\)

a: \(=\dfrac{7}{5}x^4z^3y=\dfrac{7}{5}\cdot2^4\cdot\left(-1\right)^3\cdot\dfrac{1}{2}=-\dfrac{56}{5}\)

b: \(=-xy^3\)

2) \(B=\left(1-2-3+4\right)+\left(5-6-7+8\right)+...+\left(1989-1990-1991+1992\right)+1993-1994\)

\(=0+0+...+0+1993-1994=0+1993-1994=-1\)