(x+1)(x-5)+3=0 . Tìm x

Question

(x+1)(x-5)+3=0 . Tìm x

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Mít Mít 4 years 2020-10-14T19:51:42+00:00 2 Answers 104 views 0

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  1. Nem
    0
    2020-10-14T19:52:48+00:00
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    $Ta có$ : 

    `(x + 1)(x – 5) + 3 = 0`

    $Đặt$ `t=x – 2`

    `=> (t + 3)(t – 3) + 3 = 0`

    `=> t^2 – 9 + 3 = 0` ( $hằng$ $đẳng$ $thức$ )

    `=> t^2 – 6 = 0`

    `=> (x – 2)^2 – 6 = 0`

    `=> (x – 2)^2 = 6`

    `=> x = ± √6 + 2` 

     

  2. Philomena
    0
    2020-10-14T19:53:31+00:00
    Reply

    Đáp án:

    Ta có : 

    `(x + 1)(x – 5) + 3 = 0`

    Đặt `t = x – 2`

    `=> (t + 3)(t – 3) + 3 = 0`

    `=> t^2 – 9 + 3 = 0`

    `=> t^2 – 6 = 0`

    `=> (x – 2)^2 – 6 = 0`

    `=> (x – 2)^2 = 6`

    `=> x = ± √6 + 2` 

    Đề khác :

    `(x – 1)(x – 5) + 3 = 0`

    Đặt `t = x – 3`

    `=> (t + 2)(t – 2) + 3 = 0`

    `=> t^2 – 4 + 3 = 0`

    `=> t^2 – 1 = 0`

    `=> (t – 1)(t + 1) = 0`

    `=> (x – 3 – 1)(x – 3 + 1) = 0`

    `= > (x – 4)(x – 2) = 0`

    => \(\left[ \begin{array}{l}x – 4 = 0\\x – 2 = 0\end{array} \right.\) 

    => \(\left[ \begin{array}{l}x=4\\x=2\end{array} \right.\) 

    Giải thích các bước giải:

     

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