tìm x a. 3x(x-2)-3x^2+12=0 b. x^2-9-(x-3)(2x-5)=0 c. x (3x-16)= -5

2 bình luận về “tìm x a. 3x(x-2)-3x^2+12=0 b. x^2-9-(x-3)(2x-5)=0 c. x (3x-16)= -5”

1. tìm x

   a. 3x(x – 2) – 3x² + 12 = 0

   ⇔ 3x(x – 2) – 3(x² – 4) = 0

   ⇔ 3x(x – 2) – 3(x – 2)(x + 2) = 0

   ⇔ 3(x – 2)( x – x – 2) = 0

   ⇔ -6(x – 2) = 0

   ⇔ x – 2 = 0

   ⇔ x = 2

   Vậy: x = 2

   b. x² – 9 – (x – 3)(2x – 5) = 0

   ⇔ (x – 3)(x + 3) – (x – 3)(2x – 5) = 0

   ⇔ (x – 3)(x + 3 – 2x + 5) = 0

   ⇔ (x – 3)(8 – x) = 0

   ⇔ x – 3 = 0 hoặc 8 – x = 0

   ⇔ x = 3 hoặc x = 8

   Vậy: x = 3 hoặc x = 8

   c. x(3x – 16) = -5

   ⇔ x(3x – 16) + 5 = 0

   ⇔ 3x² – 16x + 5 = 0

   ⇔ 3x² – 15x – x + 5 = 0

   ⇔ (3x² – 15x) – (x – 5) = 0

   ⇔ 3x(x – 5) – (x – 5) = 0

   ⇔ (x – 5)(3x – 1) = 0

   ⇔ x – 5 = 0 hoặc 3x – 1 = 0

   ⇔ x = 5 hoặc x = 1/3

   Vậy: x = 5 hoặc x = 1/3

   Trả lời
2. a, 3x(x – 2) – 3x^2 + 12 = 0

   => 3x^2 – 6x – 3x^2 + 12 = 0

   => -6x = 0 -12

   => -6x = -12

   => x = (-12) : (-6)

   => x =2

   Vậy x =2

   b, x^2 – 9 – (x – 3)(2x – 5) = 0

   => (x – 3)(x + 3) – (x – 3)(2x – 5) = 0

   => (x – 3)(x + 3 – 2x + 5) = 0

   => (x -3 )(8 – x) = 0

   => x – 3 = 0 hoặc 8 – x = 0

   => x = 3 hoặc x = 8

   Vậy x \in {3,8}

   c, x(3x – 16) = -5

   => 3x^2 – 16x = -5

   => 3x^2 – 16x + 5 = 0

   => 3x^2 – 15x – x + 5 = 0

   => 3x(x – 5) – (x – 5) = 0

   => (x – 5)(3x – 1) = 0

   => x – 5=  0 hoặc 3x – 1 = 0

   => x = 5 hoặc x = 1/3

   Vậy x \in {5,1/3}

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   Trả lời