a) 2/3 – 1/3 (x – 3/2) – 1/2 (2x + 1) = 5 b) 1/3x + 2/5 (x – 1) = 0

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1. a) `x(x – 2) + x – 2 = 0`

   `⇔ x(x – 2) + (x – 2) = 0`

   `⇔ (x – 2)(x + 1) = 0`

   `⇔ x – 2 = 0   hoặc   x + 1 = 0`

   * `x – 2 = 0`

   `        x = 2`

   * `x + 1 = 0`

    `      x = -1`

   Vậy pt có tập nghiệm: `S = {2; -1}`

   `b) 5x(x – 3) – x + 3 = 0`

   `⇔ 5x(x – 3) – (x – 3) = 0`

   `⇔ (x – 3)(5x – 1) = 0`

   `⇔ x – 3 = 0   hoặc   5x – 1 = 0`

   * `x – 3 = 0`

           `x = 3`

   * `5x – 1 = 0`

           5x = 1

             `x = 1/5`

   Vậy pt có tập nghiệm:

   `S = {3; 1/5}`

    

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2. a) `2/3 – 1/3 ( x – 3/2 ) – 1/2 ( 2x + 1 ) = 5`

   `=> 1/3x – 1/3 . 3/2 – 1/2 . 2x – 1/2 = 2/3 – 5 = -13/3`

   `=> 1/3x – 1/2 – 1/2 . 2x – 1/2 = -13/3`

   `=> 1/3x – 1/2 . 2x  = -13/3 + 1/2 + 1/2 = -10/3`

   `=> x . ( 1/3 – 1/2 . 2 ) = -10/3`

   `=> x . 4/3 = -10/3`

   `=> x = -5/2`

   b) `1/3x + 2/5 ( x – 1 ) = 0`

   `=> 1/3x + 2/5x – 2/5 = 0`

   `=> x . ( 1/3 + 2/5 ) = 2/5`

   `=> x . 11/15 = 2/5`

   `=> x = 2/5 : 11/15 = 6/11`

    

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