Môn Toán Lớp 8: a) (x + 4)(x – 3) + x(2 – x) = 15 b) 3x (x – 5) + 10 – 2x = 0 c) 7x (x – 11) + 2x – 22 = 0 d) (4x + 1) ² + 2x (3 – 8x) = -13 e) x(x – 7

1 bình luận về “Môn Toán Lớp 8: a) (x + 4)(x – 3) + x(2 – x) = 15 b) 3x (x – 5) + 10 – 2x = 0 c) 7x (x – 11) + 2x – 22 = 0 d) (4x + 1) ² + 2x (3 – 8x) = -13 e) x(x – 7”

1. a) (x + 4)(x – 3) + x(2 – x) = 15

   ⇔ x² + 4x – 3x – 12 + 2x – x² = 15

   ⇔ 3x = 27

   ⇔ x = 9

   Vậy: S = {9}

   b) 3x (x – 5) + 10 – 2x = 0

   ⇔ 3x (x – 5) – 2(x – 5) = 0

   ⇔ (x – 5)(3x – 2) = 0

   ⇔ x – 5 = 0 hoặc 3x – 2 = 0

   ⇔ x = 5 hoặc x = 2/3

   Vậy: S = {5; 2/3}

   c) 7x (x – 11) + 2x – 22 = 0

   ⇔ 7x (x – 11) + 2(x – 11) = 0

   ⇔ (x – 11)(7x + 2) = 0

   ⇔ x – 11 = 0 hoặc 7x + 2 = 0

   ⇔ x = 11 hoặc x = -2/7

   Vậy: S = {11; -2/7}

   d) (4x + 1) ² + 2x (3 – 8x) = -13

   ⇔ 16x² + 8+ 1 + 6x – 16x² = -13

   ⇔ 14x = 14

   ⇔ x = 1

   Vậy: S = {1}

   e) x(x – 7) – 2x + 14 = 0

   ⇔ x(x – 7) – 2(x – 7) = 0

   ⇔ (x – 7)(x – 2) = 0

   ⇔ x – 7 = 0 hoặc x – 2 = 0

   ⇔ x = 7 hoặc x = 2

   Vậy: S = {7; 2}

   f) (x + 2) ² – x(x+2) = 8

   ⇔ x² + 4x + 4 – x² – 2x = 8

   ⇔ 2x = 4

   ⇔ x = 2

   Vậy: S = {2}

   g) (4x + 1) ² – 4x + 1 = 0

   ⇔ 16x² + 8x + 1 – 4x + 1 = 0

   ⇔ 16x² + 4x + 2 = 0

   ⇔ 16x² + 8x – 4x + 2 = 0

   ⇔ x = $\frac{-1+√5}{4}$ hoặc x = $\frac{-1-√5}{4}$ 

   Vậy: S = {$\frac{-1+√5}{4}$; $\frac{-1-√5}{4}$ }

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