a: \(16^x< 32^4\)

=>\(2^{4x}< 2^{20}\)

=>4x<20

=>\(x< 5\)

=>0<=x<5

=>\(x\in\left\{0;1;2;3;4\right\}\)

b: \(9< 3^x< 81\)

=>\(3^2< 3^x< 3^4\)

=>2<x<4

=>x=3

c: \(25< 5^x< 125\)

=>\(5^2< 5^x< 5^3\)

=>2<x<3

mà x là số tự nhiên

nên \(x\in\varnothing\)

a)

`x/(-8)=5/4`

`=>x*4=5*(-8)`

`=>4x=-40`

`=>x=-40:4`

`=>x=-10`

b)

`(-13)/x=(-65)/10`

`=>(-13)/x=(-13)/2`

`=>x=2`

c)

`(-6)/5=(-12)/x`

`<=>(-12)/10=(-12)/x`

`<=>x=10`

d)

`(x-2)/2=8/(x-2)`

`=>(x-2)(x-2)=8*2`

`=>(x-2)^2=16`

`=>x-2=4` hoặc`x-2=-4`

`=>x=4+2` hoặc `x=-4+2`

`=>x=6` hoặc `x=-2`

`x/(-8) = 5/4`

`=> 4x=5.(-8)`

`=>4x=-40`

`=>x=-40:4`

`=>x=-10`

__

`-13/x =-65/10`

`=> -65x=-130`

`=>x=-130:(-65)`

`=>x= 2`

__

`-6/5 =-12/x`

`=> -6x=-12.5`

`=> -6x= -60`

`=>x=-60:(-6)`

`=>x=10`

__

`(x-2)/2 = 8/(x-2)`

`=>(x-2)^2=2.8`

`=>(x-2)^2=16`

`=> (x-2)^2=+-4^2`

\(\Rightarrow\left[{}\begin{matrix}x-2=4\\x-2=-4\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=6\\x=-2\end{matrix}\right.\)

Bài 2:

a: =>2/3x=3/4+1/2=3/4+2/4=5/4

=>x=5/4:2/3=5/4*3/2=15/8

b:=>-2x+4=3x-12

=>-5x=-16

=>x=16/5

a,3/4 . (-5/12)+3/4.(-7/12)

` 3/4 . [ - ( 5/12 + 7/12  ) ] `

`3/4 . (-1) = -3/4 `

`2/3 . x - 0,5 = 3/4 `

`        x - 0,5 = 3/4 - 2/3 `

`        x-0,5 = 1/12 `

`        x =       1/12 + 0,5 `

`        x= 7/12 `

a, 3/4 . (-5/12)+3/4.(-7/12) 

= 3/4 . [(-5/12) + (-7/12)]

= 3/4 . (-1)

= -3/4

----------------------------------------------------------------------------

a, 2/3 .x-0,5=3/4

2/3 . x = 3/2 + 0,5

2/3 . x = 2 

x = 2 : 2/3

x = 3 

vậy x = 3

a)480:x+124:x =4

(480+124):x=4

604:x=4

x=604:4

x=151

b)327:x-112:x=5

(327-112):x=5

215:x=5

x=215:5

x=43

a) 480 : x + 124 : x = 4
x : ( 480 + 124 ) = 4
x : 604 = 4
x = 2416
b) 327 : x - 112 : x = 5
x : ( 327 - 112 ) = 5
x : 215 : 5
x = 1075

a: ĐKXĐ: \(x\notin\left\{-3;2\right\}\)

b: \(A=\dfrac{x^2-4-5+x+3}{\left(x-2\right)\left(x+3\right)}=\dfrac{x^2+x-6}{\left(x-2\right)\left(x+3\right)}=\dfrac{x+2}{x-2}\)

c: Để A=3/4 thì 4x-8=3x+6

=>x=14

d: Để A nguyên thì \(x-2\in\left\{1;-1;2;-2;4;-4\right\}\)

hay \(x\in\left\{3;1;4;0;6;-2\right\}\)

\(\dfrac{25}{3}.x=\dfrac{5}{6}+\dfrac{4}{3}\)

 \(\dfrac{25}{3}.x=\dfrac{13}{6}\)

      \(x=\dfrac{13}{3}:\dfrac{25}{3}\)

     \(x=\dfrac{39}{75}\)