( x – 3 )^2 + ( x + 3 )(x + 2 ) ( 16 – 4x ) ( x + 3 ) – 4x^2 = 40 x^2 – x = 0 ( x + 1 ) ( x + 4 ) – x^2 + 3

1 bình luận về “( x – 3 )^2 + ( x + 3 )(x + 2 ) ( 16 – 4x ) ( x + 3 ) – 4x^2 = 40 x^2 – x = 0 ( x + 1 ) ( x + 4 ) – x^2 + 3”

1. a: = x^2 − 6x + 9 + x^2 + 5x + 6

   = 2x^2 − x + 15

   b: ⇔ 16x + 48 − 4 x 2 − 12x − 4x^2 = 40

   =>- 8x^2 + 4x + 48 – 40 = 0

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

   =>2x^2 – x – 2 = 0

   hay x ∈ { 1 + $\frac{1+√ 17}{4}$ ;$\frac{1-√ 17 }{4}$ }

   d: = x^2 + 5x + 4 − x^2 + 3

   = 5 x + 7

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