Tìm X: (2x +1):(2x^2 + 5x + 2) – ( 3 : x^2 -4) = 2

Question

Tìm X:
(2x +1):(2x^2 + 5x + 2) – ( 3 : x^2 -4) = 2

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Kim Cúc 9 months 2021-05-10T11:08:15+00:00 2 Answers 7 views 0

Answers ( )

    0
    2021-05-10T11:10:00+00:00

    Đáp án:

     

    Giải thích các bước giải:

    $\dfrac{\left(2x+1\right)}{2x^2+5x+2}-\:\dfrac{3}{x^2-4}=2$

    `<=>` $\dfrac{1}{x+2}-\dfrac{3}{x^2-4}=2$

    `<=>` $x-5=2\left(x+2\right)\left(x-2\right)$

    `<=>2x^2-8=x-5`

    `<=>2x^2-x-3=0`

    `<=>(2x-3)(x+1)=0`

    \(\left[ \begin{array}{l}x= \dfrac{3}{2}\\x=-1\end{array} \right.\) 

    0
    2021-05-10T11:10:13+00:00

    Đáp án:

    `S=\{-1;3/2\}`

    Giải thích các bước giải:

    `ĐKXĐ:x\ne -1/2;x\ne±2`

     `(2x+1):(2x^2+5x+2)-3:(x^2-4)=2`

    `⇔(2x+1)/(2x^2+5x+2)-3/(x^2-4)=2`

    `⇔(2x+1)/(2x^2+x+4x+2)-3/(x^2-4)=2`

    `⇔(2x+1)/[x(2x+1)+2(2x+1)]-3/(x^2-4)=2`

    `⇔(2x+1)/((2x+1)(x+2))-3/(x^2-4)=2`

    `⇔1/(x+2)-3/(x^2-4)=2`

    `⇔(x-2)/((x-2)(x+2))-3/((x-2)(x+2))=(2(x^2-4))/((x-2)(x+2))`

    `⇒(x-2)-3=2(x^2-4)`

    `⇔x-5=2x^2-8`

    `⇔2x^2-8-x+5=0`

    `⇔2x^2-x-3=0`

    `⇔2x^2+2x-3x-3=0`

    `⇔2x(x+1)-3(x+1)=0`

    `⇔(x+1)(2x-3)=0`

    \(⇔\left[ \begin{array}{l}x+1=0\\2x-3=0\end{array} \right.\)

    \(\left[ \begin{array}{l}x=-1(TM)\\x=\dfrac{3}{2}(TM)\end{array} \right.\)

    Vậy `S=\{-1;3/2\}`

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