Giải phương trình a, 4(x+2) – 1 = x-5 b, 2/x+1 – 1/x-2 = 3x -11/(x+1)(x-2) c, (x-2)(2x-1) = 5(x-2) Giải hộ tớ với ạ

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1. `a)4(x+2)-1=x-5`

   `<=>4x+8-1=x-5`

   `<=>3x=-12`

   `<=>x=-4`

   `b)2(x+1)-1/(x-2)=(3x-11)/((x+1)(x-2))`

   `=>2(x-2)-1(x+1)=3x-11`

   `<=>2x-4-x-1=3x-11`

   `<=>-2x=-6`

   `<=>x=3`

   `c)(x-2)(2x-1)=5(x-2)`

   `<=>(x-2)(2x-1)-5(x-2)=0`

   `<=>(x-2)(2x-1-5)=0`

   `<=>(x-2)(2x-6)=0`

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

   `<=>`\(\left[ \begin{array}{l}x=2\\x=3\end{array} \right.\) 

   Trả lời

2. Đáp án+Giải thích các bước giải:

    `a)4(x+2)-1=x-5`

   `⇔4x+8-1=x-5`

   `⇔4x+7=x-5`

   `⇔4x-x=-5-7`

   `⇔3x=-12`

   `⇔x=-4`

   Vậy `S=\{-4\}`

   `b)ĐKXĐ:x\ne -1;x\ne 2`

   `2/(x+1)-1/(x-2)=(3x-11)/((x+1)(x-2))`

   `⇔(2(x-2))/((x+1)(x-2))-(1(x+1))/((x-2)(x+1))=(3x-11)/((x+1)(x-2))`

   `⇒2(x-2)-(x+1)=3x-11`

   `⇔2x-4-x-1=3x-11`

   `⇔x-5=3x-11`

   `⇔3x-x=-5+11`

   `⇔2x=6`

   `⇔x=3(TM)`

   Vậy `S=\{3\}`

   `c)(x-2)(2x-1)=5(x-2)`

   `⇔(x-2)(2x-1)-5(x-2)=0`

   `⇔(x-2)[(2x-1)-5]=0`

   `⇔(x-2)(2x-1-5)=0`

   `⇔(x-2)(2x-6)=0`

   \(⇔\left[ \begin{array}{l}x-2=0\\2x-6=0\end{array} \right.\)

   \(⇔\left[ \begin{array}{l}x=2\\2x=6\end{array} \right.\)

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

   Vậy `S=\{2;3\}`

   Trả lời