`a)`

`(5-2x)^2 - 16 = 0`

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

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

`=> (1-2x)(9-2x)=0`

`=>1-2x=0` hoặc `9-2x=0`

`+) 1 - 2x = 0 => 2x = 1 => x = 1/2`

`+) 9 - 2x =0 => 2x = 9 => x = 9/2`

Vậy `x \in {1/2 ; 9/2}`

`b)`

`(x-3)^3 - (x-3)(x^2+ 3x+9) + 9(x+1)^2 = 15`

`=> (x-3) [ (x-3)^2 - (x^2 + 3x+9)] + 9 (x^2 + 2x  +1)  =15`

`=> (x-3) ( x^2-  6x + 9 - x^2 - 3x - 9) + 9x^2 + 18x + 9 = 15`

`=> (x-3) . (-9x) + 9x^2 + 18x + 9 = 15`

`=> -9x^2 + 27x + 9x^2 + 18x + 9 = 15`

`=> 45x + 9 = 15`

`=> 45x = 6`

`=> x = 2/15`

Vậy `x=2/15`

`c)`

`(x+2)(x^2 - 2x+4) - x(x^2-2) = 15`

`=> x^3 + 2^3 - x^3 + 2x =15`

`=> 2x + 8  =15`

`=> 2x = 7`

`=> x =7/2`

Vậy  `x=7/2`

`d)`

`x(3x-5)(3x+5) - (x+3)(x^2 - 3x+9) = 3 + 8x^3`

`=> x [ (3x)^2 - 5^2] - (x^3 + 3^3) =3 + 8x^3`

`=> x (9x^2 - 25) - (x^3 + 27) = 3 + 8x^3`

`=> 9x^3 - 25x - x^3 - 27 = 3 + 8x^3`

`=> 9x^3 - x^3 - 8x^3 - 25x = 3 + 27`

`=> -25x = 30`

`=> x = -6/5`

Vậy `x=-6/5`

`a)`

`(5-2x)^2 - 16 = 0`

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

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

`<=> (1-2x)(9-2x)=0`

`<=> [(1-2x=0),(9-2x=0):} <=>` \(\left[ \begin{array}{l}x=\dfrac{1}{2}\\x=\dfrac{9}{2}\end{array} \right.\) 

Vậy `x in {1/2 ; 9/2}`

`b)`

`(x-3)^3 - (x-3)(x^2+ 3x+9) + 9(x+1)^2 = 15`

`<=> (x-3)[(x-3)^2-(x^2+3x+9)]+9(x^2+2x+1)=15`

`<=>  (x-3) (x^2-6x+9-x^2-3x-9)+9x^2+18x+9= 15`

`<=> (x-3) . (-9x)+9x^2+18x+9=15`

`<=> -9x^2+27x+9x^2+18x+9=15`

`<=> 45x + 9 = 15`

`<=> 45x = 6`

`<=> x = 2/15`

Vậy `x=2/15`

`c)`

`(x+2)(x^2-2x+4)-x(x^2-2)=15`

`<=> x^3+8-x^3+2x=15`

`<=> 2x+8 =15`

`<=> 2x=7`

`<=> x = 7/2`

Vậy  `x = 7/2`

`d)`

`x(3x-5)(3x+5)-(x+3)(x^2 - 3x+9)=3+8x^3`

`<=> x (9x^2-25)-(x^3+27)=3+8x^3`

`<=> 9x^3-25x-x^3-27=3+8x^3`

`<=> 9x^3 - x^3 - 8x^3 - 25x = 3 + 27`

`<=> -25x = 30`

`<=> x = 30/-25 = -6/5`

Vậy `x=-6/5`