tìm x biết a,(2x-1)^3-4x^2(2x-3)=5 b,x^3-9x^2+27x-27=-8 c,(x+4)^3-x^2(x+12)=16 d,64x^3+48x^2+12x+1=27

0 bình luận về “tìm x biết a,(2x-1)^3-4x^2(2x-3)=5 b,x^3-9x^2+27x-27=-8 c,(x+4)^3-x^2(x+12)=16 d,64x^3+48x^2+12x+1=27”

1. Đáp án:

   a, Ta có : 

   (2x-1)^3-4x^2(2x-3)

   = 8x^3-12x^2+6x-1 – 4x^2.2x + 4x^2.3

   = 8x^3-12x^2+6x-1 – 8.x^3 + 12.x^2

   = ( 8x^3 – 8x^3) + ( 12x^2 – 12x^2 ) + 6x – 1 

   = 6x – 1 = 5 

   => 6x = 6 => x = 1

   b, Ta có : 

   ,x^3-9x^2+27x-27 = ( x – 3 )^3 = – 8 

   => x – 3 = – 2 

   => x = 1 

   c, ,(x+4)^3-x^2(x+12) 

   = x^3+12x^2+48x+64  – x^2.x – x^2.12

   = x^3+12x^2+48x+64  – x^3 – 12.x^2

   = ( x^3 – x^3)+ ( 12x^2 – 12x^2) + 48x + 64

   = 48x + 64 = 16

   => 48x = -48

   => x = -1

   d, 64x^3+48x^2+12x+1 = ( 4x+1)^3 = 27

   => 4x + 1 = 3 

   => 4x = 2 => x = 1/2

    

   Giải thích các bước giải:

    

   Trả lời

2. `a,(2x-1)^3-4x^2(2x-3)=5`

   `<=>8x^3-8x^2+2x-4x^2+4x-1-8x^3+12x^2=5`

   `<=>6x-1=5`

   `<=>x=1` 

   `b,x^3-9x^2+27x-27=-8`

   `<=>x^3-9x^2+27x-27+8=0`

   `<=>x^3-9x^2+27-19=0`

   `<=>(x^2-8x+19)(x-1)=0`

   `<=>x=1`

   `c,(x+4)^3-x^2(x+12)=16`

   `<=>x^3+8x^2+16x+4x^2+32x+64-x^3-12x^2=16`

   `<=>48x+64=16`

   `<=>48x=-48`

   `<=>x=-1`

   `d,64x^3+48x^2+12x+1=27`

   `<=>64x^3+48x^2+12x+1-27=0`

   `<=>64x^3+48x^2+12-26=0`

   `<=>32x^3+24x^2+6x-13=0`

   `<=>x=1/2`

   Vậy ….

   Trả lời