( x + 3 )^2 + ( x – 1 ) ( 2x + 5 ) x^2 + 2x – y^2 + 1 tim x ( x – 2 ) ( x + 1 ) – x^2 + 2x = 13 x^2 + 2x =0

1 bình luận về “( x + 3 )^2 + ( x – 1 ) ( 2x + 5 ) x^2 + 2x – y^2 + 1 tim x ( x – 2 ) ( x + 1 ) – x^2 + 2x = 13 x^2 + 2x =0”

1. (x + 3)^2 + (x – 1)(2x + 5)

   = x^2 + 6x + 9 + 2x^2 + 5x – 2x – 5

   = 3x^2 + 9x + 3

   —————

   x^2 + 2x – y^2 + 1

   = (x + 1)^2 – y^2

   = (x – y + 1)(x + y + 1)

   ———————-

   (x – 2)(x + 1) – x^2 + 2x = 13

   => x^2 + x – 2x – 2 – x^2 + 2x = 13

   => x – 2 = 13

   => x = 13 + 2

   => x  = 15

   Vậy x = 15

   —————–

   x^2 + 2x = 0

   => x(x + 2)  = 0

   => x = 0 hoặc x + 2 = 0

   => x = 0 hoặc x = -2

   Vậy x \in {-2,0}

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