Giải pt: $\frac{x-2}{x+2}-$$\frac{3}{x-2}$ = $\frac{2(x-11)}{x^2 – 4}$

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Giải pt:
$\frac{x-2}{x+2}-$$\frac{3}{x-2}$ = $\frac{2(x-11)}{x^2 – 4}$

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Ngọc Diệp 9 months 2021-05-07T19:57:46+00:00 2 Answers 4 views 0

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    0
    2021-05-07T19:59:33+00:00

    Đáp án:

     $\dfrac{x-2}{x+2} – \dfrac{3}{x-2} =\dfrac{2(x-11)}{x^2-4}$

    $\text{ĐKXĐ : $x \neq ± 2 $ }$

    $⇔ \dfrac{(x-2)(x-2)}{(x+2)(x-2)} – \dfrac{3(x+2)}{(x-2)(x+2)} = \dfrac{2(x-11)}{(x-2)(x+2)}$

    $⇒(x-2)^2 – 3(x+2) = 2(x-11)$

    $⇔x^2 -4x +4 -3x -6 = 2x -22$

    $⇔x^2 -4x -3x -2x +4 -6 +22 =0$

    $⇔x^2 -9x +20 =0$

    $⇔x^2 -2 . x . \dfrac{9}{2} + (\dfrac{9}{2})^2 – \dfrac{1}{4} = 0$

    $⇔ (x – \dfrac{9}{2})^2 – \dfrac{1}{4} =0$

    $⇔ (x-\dfrac{9}{2})^2 – (\dfrac{1}{2})^2 =0$

    $⇔ (x – \dfrac{9}{2} – \dfrac{1}{2} ).( x – \dfrac{9}{2} + \dfrac{1}{2}) =0$

    $⇔ $\(\left[ \begin{array}{l}x-\dfrac{9}{2} – \dfrac{1}{2}=0\\x-\dfrac{9}{2} +\dfrac{1}{2}=0\end{array} \right.\) 

    $⇔$\(\left[ \begin{array}{l}x=5 (TM)\\x=4 (TM)\end{array} \right.\) 

    $\text{Vậy phương trình có tập nghiệm S ={5 ; 4 } }$

     

    0
    2021-05-07T19:59:38+00:00

    Đáp án:

    \(\rm S=\left\{{4;5}\right\}\)

    Giải thích các bước giải:

    \(\dfrac{x-2}{x+2}-\dfrac{3}{x-2}=\dfrac{2(x-11)}{x^2-4}\,\,\,\,\,ĐKXĐ:\begin{cases}x\neq2\\x\neq-2\end{cases}\\\Leftrightarrow\dfrac{(x-2)^2-3(x+2)}{(x+2)(x-2)}-\dfrac{2(x-11)}{(x+2)(x-2)}=0\\\Rightarrow(x-2)^2-3(x+2)-2(x-11)=0\\\Leftrightarrow x^2-4x+4-3x-6-2x+22=0\\\Leftrightarrow x^2-9x+20=0\\\Leftrightarrow x^2-4x-5x+20=0\\\Leftrightarrow x(x-4)-5(x-4)=0\\\Leftrightarrow(x-4)(x-5)=0\\\Leftrightarrow\left[ \begin{array}{l}x-4=0\\x-5=0\end{array} \right.\\\Leftrightarrow\left[ \begin{array}{l}x=4\\x=5\end{array} \right.\rm (tmđk)\\Vậy\ S=\left\{{4;5}\right\}\)

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