a, (x-1)³ -(x+3).(x² -3x+9)+3(x²-4)=2
<=>x³ - 3x² + 3x - 1 - x³ - 27 + 3x² - 12 - 2 = 0
<=>3x = 42
<=> x = 14 Vậy x = 14
b) (x+2) (x² -2x+4)- x (x+2)²=0
<=> x³ + 8 - x ( x² + 4x + 4) = 0
<=> x³ + 8 - x³ - 4x² - 4x = 0
<=> -4x² - 4x + 8 = 0
<=> -4 ( x² + x - 2 ) = 0
<=> x² + x - 2 = 0
<=> x² - x + 2x - 2 = 0
<=> x ( x-1) + 2 (x-1) = 0
<=> (x-1) ( x+2) = 0
<=> - TH1: x-1 = 0 --> x=1
- TH2 : x+2 = 0 --> x = -2
Vậy x ∈ {1;2}
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