Đáp án:
ĐKXĐ : ` x \ne 2`
` (2(x^2+x+6))/(x^3-8) + 2/(2-x) = 3/(x^2+2x+4)`
` => (2x^2+2x+12)/((x-2)(x^2+2x+4)) - (2(x^2+2x+4))/((x-2)(x^2+2x+4)) = (3(x-2))/((x-2)(x^2+2x+4))`
` => (2x^2+2x+12)/((x-2)(x^2+2x+4)) - (2(x^2+2x+4))/((x-2)(x^2+2x+4)) - (3(x-2))/((x-2)(x^2+2x+4)) = 0`
` => (2x^2+2x +12 - 2x^2 -4x -8 - 3x + 6)/((x-2)(x^2+2x+4)) = 0`
` => 2x^2+2x +12 - 2x^2 -4x -8 - 3x + 6 = 0`
` => -5x +10=0`
` => -5x = -10`
` => x= 2`
Mà ` x \ne 2` nên PTVN
