Giải bất phương trình:(5x-3/x+2)-4=6/x+2

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1. $\dfrac{5x-3}{x+2}-4=\dfrac{6}{x+2}$

   $↔\dfrac{5x-3}{x+2}=\dfrac{6}{x+2}+4$

   $↔\dfrac{5x-3}{x+2}=\dfrac{14+4x}{x+2}$

   $↔\dfrac{5x-3}{x+2}-\dfrac{14+4x}{x+2}=0$

   $↔\dfrac{5x-3-(14+4x)}{x+2}=0$

   $↔\dfrac{x-17}{x+2}=0$

   $↔\dfrac{x}{x+2}-\dfrac{17}{x+2}=0$

   $↔\dfrac{x}{x+2}=\dfrac{17}{x+2}$

   $↔x(x+2)=17(x+2)$

   $↔x=17$

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2. Đáp án:

    `(5x-3)/(x+2)-4=6/(x+2)`

   `⇔(5x-3)/(x+2)-6/(x+2)=4`

   `⇔(5x-3-6)/(x+2)=4`

   `⇔(5x-9)/(x+2)=4`

   `⇔4(x+2)=5x-9`

   `⇔4x+8=5x-9`

   `⇔5x-4x=8+9`

   `⇔x=17`

   vậy `x=17`

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