tìm x biết a) $x^{3}$ + 2x – 3 =0 b) $x^{3}$ – $9x^{2}$ +6x + 16=0 c) $x^{3}$ – 2x -4=0

1 bình luận về “tìm x biết a) $x^{3}$ + 2x – 3 =0 b) $x^{3}$ – $9x^{2}$ +6x + 16=0 c) $x^{3}$ – 2x -4=0”

1. $#duong612009$

   a, x^3 + 2x – 3= 0

   => x^3 – x + 3x – 3= 0

   => x(x^2 – 1) + 3(x – 1) = 0

   => x(x – 1)(x + 1) + 3(x – 1) = 0

   => (x – 1)[x(x + 1) + 3] = 0

   => (x – 1)(x^2 + x + 3) = 0

   Vì x^2 + x + 3 = x^2 + 2.x.1/2 + (1/2)^2 + (11)/4 = (x + 1/2)^2 + (11)/4 > 0

   => x – 1=  0

   => x = 1

   Vậy x =1

   b, x^3 – 9x^2 + 6x + 16 = 0

   => x^3 – 8x^2 – x^2 + 8x – 2x + 16 = 0

   => x^2 (x – 8) – x(x – 8) – 2(x – 8) = 0

   => (x – 8)(x^2 – x – 2) = 0

   => (x – 8)(x^2  – 2x + x – 2) = 0

   => (x – 8)[x(x – 2) + (x – 1)] = 0

   => (x – 8)(x – 2)(x + 1) = 0

   => x – 8 = 0 hoặc x – 2 =0 hoặc x + 1 = 0

   => x = 8 hoặc x = 2 hoặc x = -1

   Vậy x \in {-1,2,8}

   c, x^3 – 2x – 4 = 0

   => x^3 – 2x^2 + 2x^2 – 4x + 2x – 4 = 0

   => x^2 (x – 2) + 2x(x – 2) + 2(x – 2) = 0

   => (x – 2)(x^2 + 2x + 2) = 0

   Vì x^2 + 2x + 2 =  x^2 + 2x + 1 + 1 = (x + 1)^2 + 1 > 0

   => x – 2 = 0

   => x = 2

   Vậy x = 2

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