Giải thích các bước giải:
`x^4 - x^3 + x-1 = (x-1) (x^3 + 1) = (x-1)(x+1) (x^2-x+1)`
`x^4 + x^3 - x -1 = (x+1) (x^3 -1) = (x+1)(x-1)(x^2 + x+1)`
`x^5 - x^4 + x^3 - x^2 + x+1 = (x+1)(x^4 + x^2 + 1)`
Ta có:
`3/(x^4 - x^3 + x-1) - 1/ (x^4 + x^3 -x -1)`
`= (3 (x^2 + x+1) - (x^2 - x + 1))/ ((x-1)(x+1)(x^2-x+1)(x^2+x+1))`
`= (2x^2 + 4x + 4) / ((x-1)(x+1)(x^4 + x^2 +1))`
Vậy `P = (2x^2 + 4x + 2)/((x-1)(x+1)(x^4 + x^2)) - 4/((x-1)(x^4 + x^2 + 1))`
`= (2x^2 + 4x + 2 -4(x+1))/ ((x-1)(x+1)(x^4+x^2+1)) = (2x^2 -2)/((x-1)(x+1)(x^4 + x^2 +1))`
`= (2(x-1)(x+1))/ ((x-1)(x+1)(x^4 + x^2 +1)) = 2/ (x^4 + x^2 +1)` `(x ne ± 1)`
Do `x^4 + x^2 + 1 = (x^2 + 1/2)^2 + 3/4 > 0 ∀ x`
Vậy `P > 0 ∀ x ne ± 1`
