`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}`
