\(x-1,2=18,6:3\)

\(x-1,2=6,2\)

\(x=6,2+1,2\)

\(x=7,4\)

vậy \(x=7,4\)

những câu khác tương tự 

a,x - 1,44 : 1,2 = 5,9

x-1,2=5,9

x=5,9+1,2

x=7,1

b,86,5 - x : 8,6 = 23,7

x:8,6=86,5-23,7

x:8,6=62,8

x=62,8 x 8,6

x=540,08

c,( 25,34 - x ) : 2,8 - 1,2 = 3,7

( 25,34 - x ) : 2,8=3,7+1,2

( 25,34 - x ) : 2,8=4,9

( 25,34 - x ) =4,9 x 2,8

( 25,34 - x ) =13,72

 x = 25,34-13,72

x=11,62

d,x + 75% x - 0,25 x = 22,5

x + 0,75 x-0,25x=22,5

x (1+0,75-0,25)=22,5

x * 1,5 = 22,5

x= 22,5:1,5

x=15

e,( x + 1 ) + ( x + 5 ) + ( x +  9 ) + ... + ( x + 29 ) = 960

(x+x+x+x+x+x+x+x)+(1+5+9+13+17+21+25+29)=960

8*x+120=960

8*x=960-120

8*x=840

x=840:8

x=105

a)

` x - 1,44 : 1,2 = 5,9`

` x - 1,2 = 5,9`

`x = 5,9 + 1,2`

`x = 7,1`

b)

` 86,5 - x : 8,6 = 23,7`

` x : 8,6 = 86,5 - 23,7`

` x : 8,6 =  62,8`

` x = 62,8 xx 8,6`

`x = 540,08`

c)

`( 25,34 - x ) : 2,8 - 1,2 = 3,7`

`( 25,34 - x ) : 2,8 =  3,7 + 1,2`

`( 25,34 - x ) : 2,8 = 4,9`

` 25,34 - x = 4,9 xx 2,8`

` 25,34 -x = 13,72`

` x = 25,34 - 13,72`

`x = 11,62`

e)

` ( x + 1 ) + ( x + 5 ) + ( x +  9 ) + ... + ( x + 29 ) = 960`

Có tất cả số hạng là:

\(\left(29-1\right):4+1=8\left(số\right)\)

Tổng dãy số :

\(\left(29+1\right)\times8:2=120\)

` => ( x + x + x+....+ x) + ( 1 + 5 + 9 +....+ 29) = 960`

` 8x + 120 = 960`

` 8x = 960 - 120`

`8x = 840`

` x = 840 : 8`

`x = 105`

mn ơi giúp mik ik mik cần gấp

a)

\(\begin{array}{l}{(1,2)^3}.x = {(1,2)^5}\\x = {(1,2)^5}:{(1,2)^3}\\x = {(1,2)^2}\\x = 1,44\end{array}\)

Vậy \(x = 1,44\).

b)

\(\begin{array}{l}{\left( {\frac{2}{3}} \right)^7}:x = {\left( {\frac{2}{3}} \right)^6}\\x = {\left( {\frac{2}{3}} \right)^7}:{\left( {\frac{2}{3}} \right)^6}\\x = \frac{2}{3}\end{array}\)

Vậy \(x = \frac{2}{3}\).

a: =>1,2<x<2,4

=>x=2

b: =>2,76<x<3,51

=>x=3

a: y=8

bạn vào trang cá nhân tặng coin cho mik rùi mik giải ch

a) Ta có: \(-1\dfrac{1}{3}-x=2.5\)

\(\Leftrightarrow\dfrac{-4}{3}-x=\dfrac{5}{2}\)

hay \(x=\dfrac{-4}{3}-\dfrac{5}{2}=\dfrac{-8}{6}-\dfrac{15}{6}=\dfrac{-23}{6}\)

b) Ta có: \(\dfrac{1}{3}:x-20\%=\dfrac{16}{-32}\)

\(\Leftrightarrow\dfrac{1}{3}:x=\dfrac{-1}{2}+\dfrac{1}{5}=\dfrac{-3}{10}\)

hay \(x=\dfrac{1}{3}:\dfrac{-3}{10}=\dfrac{1}{3}\cdot\dfrac{10}{-3}=\dfrac{-10}{9}\)

c) Ta có: \(\left(\dfrac{3x}{7}+1\right):\left(-4\right)=\dfrac{-1}{28}\)

\(\Leftrightarrow\dfrac{3}{7}x+1=\dfrac{1}{7}\)

\(\Leftrightarrow x\cdot\dfrac{3}{7}=\dfrac{1}{7}-1=\dfrac{-6}{7}\)

hay x=-2

a) Ta có: ⇔−43−x=52⇔−43−x=52

hay 13:x−20%=16−3213:x−20%=16−32

x=13:−310=13⋅10−3=−109x=13:−310=13⋅10−3=−109

c) Ta có: ⇔37x+1=17⇔37x+1=17

Trả lời :

a, 295,1 : ( x - 13 ) = 13

=> x - 13 = 295,1 : 13

x - 13 = 22,7

=> x = 22,7 + 13

x =  35,7

b, ( 8,2 + x ) . 4,1 = 184,5

=> 8,2 + x = 184,5 : 4,1

8,2 + x = 45

=> x = 45 - 8,2

x = 36,8

c, ( x + 1,2 ) x 26,7 = 961,2

=> x + 1,2 = 961,2 : 26,7

x + 1,2 = 36

=> x = 36 - 1,2

x = 34,8

a, 295,1 : ( x - 13 ) = 13

x - 13 = 295,1 : 13

x - 13 = 22,7

x = 22,7 + 13

x = 45,7

b, ( 8,2 + x ) x 4,1 = 184,5

8,2 + x = 184,5 : 4,1

8,2 + x = 45

x = 45 - 8,2

x = 36,8

c, ( x + 1,2 ) x 26,7 = 961,2

x + 1,2 = 961,2 : 26,7

x + 1,2 = 36

x = 36 - 1,2

x = 34,8

a,295,1:(x-13)=13

              x-13=295,1:13

              x-13=22,7

                   x=22,7+13

                   x=35,7

b,(8,2+x)x4,1=184,5

            8,2+x=184,5:4,1

            8,2+x=45

                   x=45-8,2

                   x=36,8

c,(x+1,2)x26,7=961,2

              x+1,2=961,2:26,7

              x+1,2=36

                     x=36-1,2

                     x=34,8

a) Ta có số đối của 2,5 là -2,5

\(\Rightarrow x-3,5=-2,5\)

\(\Rightarrow x=-2,5+3,5\)

\(\Rightarrow x=1\)

b) Ta có số đối của -12 là 12

\(\Rightarrow3x-12=12\)

\(\Rightarrow3x=24\)

\(\Rightarrow x=\dfrac{24}{3}=8\)

c) Ta có số đối của \(-\dfrac{1}{8}\) là \(\dfrac{1}{8}\)

\(\Rightarrow2x+\dfrac{1}{4}=\dfrac{1}{8}\)

\(\Rightarrow2x=\dfrac{1}{8}-\dfrac{1}{4}\)

\(\Rightarrow2x=-\dfrac{1}{8}\)

\(\Rightarrow x=-\dfrac{1}{8}:2\)

\(\Rightarrow x=-\dfrac{1}{16}\)

d) Bạn viết lại đề

a, \(\left|2x-3\right|-\dfrac{1}{3}=0\Leftrightarrow\left|2x-3\right|=\dfrac{1}{3}\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-3=\dfrac{1}{3}\\2x-3=-\dfrac{1}{3}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{3}\\x=\dfrac{4}{3}\end{matrix}\right.\)

b, tương tự 

c, \(\left|2x-1\right|-\left|x+\dfrac{1}{3}\right|=0\Leftrightarrow\left|2x-1\right|=\left|x+\dfrac{1}{3}\right|\)

TH1 : \(2x-1=x+\dfrac{1}{3}\Leftrightarrow x=\dfrac{4}{3}\)

TH2 : \(2x-1=-x-\dfrac{1}{3}\Leftrightarrow3x=\dfrac{2}{3}\Leftrightarrow x=\dfrac{2}{9}\)

d, \(3x-\left|x+15\right|=\dfrac{5}{4}\Leftrightarrow\left|x+15\right|=3x-\dfrac{5}{4}\)ĐK : x >= 5/12

TH1 : \(x+15=3x-\dfrac{5}{4}\Leftrightarrow-2x=-\dfrac{65}{4}\Leftrightarrow x=\dfrac{65}{8}\)( tm )

TH2 : \(x+15=\dfrac{5}{3}-3x\Leftrightarrow4x=-\dfrac{40}{3}\Leftrightarrow x=-\dfrac{10}{3}\)

b: ta có: \(\dfrac{5}{6}-\left|x+\dfrac{1}{4}\right|=\dfrac{1}{4}\)

\(\Leftrightarrow\left|x+\dfrac{1}{4}\right|=\dfrac{7}{12}\)

\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{4}=\dfrac{7}{12}\\x+\dfrac{1}{4}=-\dfrac{7}{12}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{3}\\x=-\dfrac{5}{6}\end{matrix}\right.\)