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

Question

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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Calantha 1 year 2020-11-26T18:00:11+00:00 2 Answers 151 views 0

Answers ( )

  1. Euphemia
    0
    2020-11-26T18:01:28+00:00
    Reply

    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`

     

  2. dangkhoa
    0
    2020-11-26T18:01:49+00:00
    Reply

    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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Giải phương trình 1 ẩn: x + 2 - 2(x + 1) = -x . Hỏi x = ? ( )