a) (x - 1)3 + (2 - x)(4 + 2x + x2) + 3x(x + 2) = 16
x3 - 3x2 + 3x - 1 + 8 - x3 + 3x2 + 6x - 16 = 0
9x - 9 = 0
9x = 9
x = 1
Vậy x ∈ {1}
b) ( x + 2)(x2 - 2x + 4) - x(x2 - 2) = 16
x3 + 8 - x3 + 2x - 16 = 0
2x - 8 = 0
2x = 8
x = 4
Vậy x ∈ {4}
c) x(x - 5)(x + 5) - (x + 2)(x2 - 2x + 4) = 17
x3 - 25x - x3 - 8 - 17 = 0
-25x - 25 = 0
-25x = 25
x = -1
Vậy x ∈ {1}
d) (x - 3)3 - (x - 3)(x2 + 3x + 9) + 9(x + 1)2 = 15
x3 - 9x2 + 27x - 27 - x3 + 27 + 9x2 + 18x + 9 - 15 = 0
45x - 6 = 0
45x = 6
x = \(\frac{2}{15}\)
Vậy x ∈ {\(\frac{2}{15}\)}