Tìm x a,(x-1)^3 -(x+3).(x^2 -3x+9)+3(x^2-4)=2 b,(x+2)(x^2 -2x+4)-x(x+2)^2=0

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1. a, `(x-1)^3-(x+3)(x^2-3x+9)+3(x^2-4)=2`

   `<=> x^3-3x^2+3x-1-x^3-27+3x^2-12=2`

   `<=> 3x-40=2`

   `<=> 3x=42`

   `<=> x=14`

   b, `(x+2)(x^2-2x+4)-x(x+2)^2=0`

   `<=> x^3+8-x(x^2+4x+4)=0`

   `<=> x^3+8-x^3-4x^2-4x=0`

   `<=> -4x^2-4x+8=0`

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

   `<=> x^2+2x-x-2=0`

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

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

   `<=> x in {1;-2}`

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2. 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}

   NHỚ VOTE 5 SAO VÀ CTLHN CHO MIK NHÉ. THANKS!!!

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