40 . x = 100 . ( x - 18 )

        40 x   = 100 x - 1800

40x - 100x = -1800

  - 60 x       = - 1800

         x       = 30

        40 . x = 100 . ( x - 18 )

        40 x   = 100 x - 1800

40x - 100x = -1800

  - 60 x       = - 1800

         x       = 30

40 .x = 100 . ( x - 18)

40 .x = 100 x  -1800

40x - 100 x= - 1800

- 60 x       =- 1800

x              =30

a; 100% : \(x\) - 50% : \(x\) + 40% : \(x\) = 18 + 30% : \(x\)

   (100% - 50% + 40%): \(x\) = 18 + 30% : \(x\)

                            90% : \(x\)  = 18 + 30% : \(x\)                  

                           90%: \(x\)  - 30%: \(x\) = 18

                          60% : \(x\) = 18

                                   \(x\) = 60% : 18

                                    \(x\) = \(\dfrac{1}{30}\)

100%:x-50%:x+40%:x=18+30%:x

<=> (100% - 50% + 40%) : x = 18 + 30% : x

<=> 90% : x = 18 + 30% : x 

=> 90% : x - 30% : x = 18

=> (90% - 30%) : x = 18

<=> x = 60% : 18

=> x = \(\frac{1}{30}\)

80 + X = 100 + 40

80 + X = 140

       X = 140 - 80

       X = 60

Chúc bạn may mắn

60 nhé bạn

Tk nha kb nữa☺☺☺

80+x=140

x=140-80

x=60

k cho minh nha

`@` `\text {Ans}`

`\downarrow`

`[ (100 - x)*2 - 2] = 40`

`=> (100-x)*2 - 2 = 40`

`=> (100 - x)*2 = 40 + 2`

`=> (100 - x)*2 = 42`

`=> 100 - x = 42 \div 2`

`=> 100 - x = 21`

`=> x = 100 - 21`

`=> x = 79`

Vậy, `x=79.`

\(\left[\left(100-x\right).2-2\right]=40\)

\(=>\left(100-x\right).2=40+2\)

\(=>\left(100-x\right).2=42\)

\(=>100-x=42:2\)

\(=>100-x=21\)

\(=>x=100-21\)

\(=>x=79\)

Chúc bạn học tốt

[ ( 100 - x ) . 2 - 2 ] = 40

<=> [ (200 - 2x) -2] = 40

<=> 198 - 2x = 40

<=> -2x = -158

<=> x = 79

|50-x|=40

=>50-x=40                  hoặc              50-x=-40

         x=50-40                                        x=50-(-40)

         x=10                                             x=90

|100-x+5|=-7

vô lí

tìm x biết 

100% : x - 50% : x + 40 % : x = 18 + 30% 

1 : x - 0.5 + 0.4 : x = 18 + 0.3

( 1 - 0.5 + 0.4 ) : x = 17.7

0.9 : x = 17.7

       x = 0.9 : 17.7

      x = \(\frac{3}{59}\)

Theo bài ra ta có : x < 6

\(\hept{\begin{cases}40:x=dư6\\100:x=dư4\end{cases}}=>\hept{\begin{cases}40-6⋮x\\100-4⋮x\end{cases}}=>\hept{\begin{cases}34⋮x\\96⋮x\end{cases}}\)

=> x \(\in\)ƯC_(34;96)

mà 34 = 17 . 2

     96  = 2⁵ . 3 

=> ƯCLN (34,96) = 2

Lại có ƯC (34,96) = Ư(2) \(\in\){1;2}

=> \(\hept{\begin{cases}x=1\\x=2\end{cases}}=>x\in\varnothing\)