Giải các phương trình sau: a) (3x + 1) (x – 3)^2 = (3x + 1) (2x – 5)^2 b) 3/5x – 1 + 2/ 3 – 5x = 4/ (1 – 5x) (5x – 3) c) ║5x-1║ – 2x = 7

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1. `a)(3x+1)(x-3)²=(3x+1)(2x-5)²`

   `⇔(3x+1)(x-3)²-(3x+1)(2x-5)²=0`

   `⇔(3x+1)[(x-3)²-(2x-5)²]=0`

   `⇔(3x+1)(x-3+2x-5)(x-3-2x+5)=0`

   `⇔(3x+1)(3x-8)(-x+2)=0`

   `⇔`\(\left[ \begin{array}{l}3x+1=0\\3x-8=0\\-x+2=0\end{array} \right.\)

   `⇔`\(\left[ \begin{array}{l}x=\dfrac{-1}{3}\\x=\dfrac{8}{3}\\x=2\end{array} \right.\)

   Vậy `S={(-1)/3;8/3;2}`

   `b)3/(5x-1)+2/(3-5x)=4/[(1-5x)(5x-3)“(“ĐKXĐ:“x`$\neq$ `1/5,x`$\neq$ `3/5)`

   `⇔3/(5x-1)+2/(3-5x)=4/[(5x-1)(3-5x)]`

   `⇔[3(3-5x)]/[(5x-1)(3-5x)]+[2(5x-1)]/[(5x-1)(3-5x)]=4/[(5x-1)(3-5x)]`

   `⇒3(3-5x)+2(5x-1)=4`

   `⇔9-15x+10x-2=4`

   `⇔-5x+7=4`

   `⇔-5x=4-7`

   `⇔-5x=-3`

   `⇔x=3/5(loại)`

   Vậy `S=∅`

   `c)|5x-1|-2x=7`

       `|5x-1|=5x-1` nếu `x≥1/5`

                    `-5x+1` nếu `x<1/5`

   Nếu `x≥1/5` ta có:

        `5x-1-2x=7`

   `⇔3x-1=7`

   `⇔3x=7+1`

   `⇔3x=8`

   `⇔x=8/3(TM)`

   Nếu `x<1/5` ta có:

        `-5x+1-2x=7`

   `⇔-7x-1=7`

   `⇔-7x=7-1`

   `⇔-7x=6`

   `⇔x=-6/7(TM)`

   Vậy `S={8/3;-6/7}`

    

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