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

`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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