giúp tôi nhá

a) 5(x+7) - 10 = 2^3 . 5

 5(x+7 ) -10 = 8 . 5 = 40

5(x+7) = 40 + 10 = 50

x + 7 = 50 : 5 = 10

x = 10 - 7 = 3

b) 2^6 + ( 5 + x ) = 3^4

  64 +  ( 5 + x ) = 81

5 + x = 81 - 64 = 17

x = 17 - 5 = 12

a) 2x - 20 = x + 30

2x - x = 30 + 20

x = 50

b) 5x - 30 + 4x = 60

9x = 60 + 30

9x = 90

x = 90 : 9

x = 10

a) 2x - 20 = x + 30

2x - x = 30 + 20

x = 50

b) 5x - 30 + 4x = 60

9x = 60 + 30

9x = 90

x = 90 : 9

x = 10

\(2.x-20=x+30\)

\(2.x=x+30-20\)

\(2.x=x+10\)

\(x=10\)

Tách ?

`a, 70 -5.(x-3) =45`

`=> 5.(x-3)= 70-45`

`=> 5.(x-3)=25`

`=>x-3=25:5`

`=>x-3=5`

`=>x= 5+3`

`=>x=8`

______

`b,10 + 2.x = 4^5:4^3`

`=> 10 + 2.x = 4^(5-3)`

`=> 10 + 2.x =4^2=16`

`=> 2.x=16-10`

`=>2.x=6`

`=>x=6:2`

`=>x=3`

_____

`c,60-3.x-2=51`

`=>  60-3.x= 51+2`

`=> 60-3.x=53`

`=>3.x=60-53`

`=> 3.x= 7`

`=>x= 7/3`

____

`d, 4.x-20=2^5:2^3`

`=> 4.x-20=2^(5-3)`

`=> 4.x-20=2^2`

`=> 4.x= 4+20`

`=>4.x=24`

`=>x=24:4`

`=>x=6`

____

`2^x . 4=16`

`=> 2^x=16:4`

`=>2^x= 4`

`=>2^x=2^2`

`=>x=2`

____

`f, 3^x . 3=243`

`=>3^x=243:3`

`=> 3^x=81`

`=> 3^x=3^3`

`=>x=3`

_____

`g, 64. 4^x =16^8`

`=> 4^3 . 4^x=(4^2)^8`

`=> 4^3 . 4^x = 4^(16)`

`=> 4^x= 4^(16-3)`

`=>4^x=4^(13)`

`=>x=13`

_____

`2^x . 16^2 =1024`

`=> 2^x= 1024 : 16^2`

`=>2^x=4`

`=>2^x=2^2`

`=>x=2`

a: =>5(x-3)=25

=>x-3=5

=>x=8

b: =>2x=16-10=6

=>x=3

c: =>58-3x=51

=>3x=7

=>x=7/3

d: =>4x-20=4

=>4x=24

=>x=6

e: =>2^x=4

=>2^x=2^2

=>x=2

f: =>3^x=81

=>3^x=3^4

=>x=4

g: =>4^x*4^3=4^16

=>x+3=16

=>x=13

h: =>2^x=1024/256=4=2^2

=>x=2

b: =>5x=48,5

hay x=9,7

X − 17 = 47

X = 47 + 17

X = 64

X + 26 = 60

X = 60 − 26

X = 34

38 + x = 54

X = 54 − 38

X = 16

d) Ta có: \(32\%-0.25:x=-\dfrac{17}{5}\)

\(\Leftrightarrow0.25:x=\dfrac{8}{25}+\dfrac{17}{5}=\dfrac{93}{25}\)

hay \(x=\dfrac{25}{372}\)

Vậy: \(x=\dfrac{25}{372}\)

e) Ta có: \(\left(x+\dfrac{1}{5}\right)^2+\dfrac{17}{25}=\dfrac{26}{25}\)

\(\Leftrightarrow\left(x+\dfrac{1}{5}\right)^2=\dfrac{9}{25}\)

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

Vậy: \(x\in\left\{\dfrac{2}{5};-\dfrac{4}{5}\right\}\)

f) Ta có: \(-\dfrac{32}{27}-\left(3x-\dfrac{7}{9}\right)^3=-\dfrac{24}{27}\)

\(\Leftrightarrow\left(3x-\dfrac{7}{9}\right)^3=\dfrac{-8}{27}\)

\(\Leftrightarrow3x-\dfrac{7}{9}=-\dfrac{2}{3}\)

\(\Leftrightarrow3x=\dfrac{1}{9}\)

hay \(x=\dfrac{1}{27}\)

g) Ta có: \(60\%\cdot x+0.4x+x:3=2\)

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

hay \(x=\dfrac{3}{2}\)

Vậy: \(x=\dfrac{3}{2}\)

h) PT \(\Leftrightarrow\left|\dfrac{20}{9}-x\right|=\dfrac{2}{9}\) \(\Rightarrow\left[{}\begin{matrix}\dfrac{20}{9}-x=\dfrac{2}{9}\\x-\dfrac{20}{9}=\dfrac{2}{9}\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{22}{9}\end{matrix}\right.\)

  Vậy ...

i) PT \(\Leftrightarrow\dfrac{8}{5}+\dfrac{2}{5}x=\dfrac{16}{5}\) \(\Leftrightarrow\dfrac{2}{5}x=\dfrac{8}{5}\) \(\Leftrightarrow x=4\)

  Vậy ...

b. \(\dfrac{13+x}{20}\)= \(\dfrac{15}{20}\)

=>  13+x=15

            x=15-13

            x=2

$#Shả$

`x xx1,2+x xx1,8=45`

`<=>x xx(1,2+1,8)=45`

`<=> x xx 3 =45`

`<=>x=45:3=15`

`(13+x)/20=3/4`

`<=>4xx(13+x)=20xx3`

`<=>4xx(13+x)=60`

`<=>13+x=60:4=15`

`<=>x=15-13=2`

a. x x(1,2+1,8)=45

    x x 3= 45

    x= 45:3

    x=15

a: \(x\cdot3\cdot5=2,7\)

=>\(x\cdot15=2,7\)

=>\(x=\dfrac{2,7}{15}=0,18\)

b: \(1,2:x\cdot4=20\)

=>\(1,2:x=20:4=5\)

=>\(x=1,2:5=0,24\)

a) \(\dfrac{x}{-6}=\dfrac{-15}{45}\)

    \(\dfrac{-x}{6}=\dfrac{-15}{45}\)

    \(\dfrac{x}{6}=\dfrac{15}{45}\)

    \(x=\dfrac{\left(15\cdot6\right)}{45}\)

    \(x=2\)

b) \(\dfrac{x}{5}=\dfrac{16}{25}\)

    \(x=\dfrac{\left(16\cdot5\right)}{25}\)

    \(x=\dfrac{80}{25}\)

   \(x=\dfrac{16}{5}\)

c) \(\dfrac{5}{x-3}=\dfrac{20}{-12}\)

  \(x-3=\dfrac{\left(5\cdot-12\right)}{20}\)

 \(x-3=-3\)

 \(x=\left(-3\right)+3\)

\(x=0\)

d) \(\dfrac{2}{5}\cdot x=\dfrac{6}{35}\)

         \(x=\dfrac{6}{35}\div\dfrac{2}{5}\)

        \(x=\dfrac{3}{7}\)