Question

bài 6 tìm x 

a) 8x³ - 12x² + 6x - 1 = 0                                  b) x³ - 6x² + 12x - 8 = 27

c) x² - 8x + 16 = 5 ( 4 - x ) ³                               d) (2-x)³ = 6x ( x - 2 )

Answer 1

`a, 8x^3 - 12x^2 + 6x - 1 = 0`

`=> (2x - 1)^3 = 0`

`=> 2x - 1 =0`

`=> 2x = 1`

`=>x = 1:2`

`=>x=1/2`

Vậy: `x=1/2`

`b, x^3 - 6x^2 + 12x - 8 = 27`

`=> (x - 2)^3= 27`

`=> (x-2)^3 = 3^3`

`=>x-2=3`

`=>x=3+2`

`=>x=5`

Vậy: `x=5`

 

Answer 2

`c, x^2 - 8x + 16 = 5(4-x)^3`

`=> (x-4)^2 -5(4-x)^3=0`

`=> (x-4)^2 + 5(x-4)^3=0`

`=> (x-4)^2[1+5(x-4)]= 0`

`=> (x-4)^2(5x-3)=0`

`=> [((x-4)^2 = 0),(5x-3=0):}`

`=> [(x = 4),(x = 3/5):}`

Vậy: `x=4;x=3/5`

`d, (2-x)^3 = 6x(x-2)`

`=> (2-x)^3 - 6x(x-2)=0`

`=> (2-x)^3 +6x(2-x)=0`

`=> (2-x)[(2-x)^2 + 6x]=0`

`=> (2-x)(4-4x+x^2 + 6x)=0`

`=> (2-x)(x^2 + 2x +4)=0`

`=> [(2-x=0),(x^2 + 2x + 4 \ne 0):}`

`=> x=2`

Vậy: `x=2`