`7)(3x-5)^2-(x-4)^2=0`
`<=>(3x-5-x+4)(3x-5+x-4)=0`
`<=>(2x-1)(4x-9)=0`
`<=>` $\left[\begin{matrix} x=\dfrac{1}{2}\\ x=\dfrac{9}{4}\end{matrix}\right.$
Vậy `S={1/2;9/4}`
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`8)(x^2-6x+9)-(5-2x)^2=0`
`<=>(x-3)^2-(5-2x)^2=0`
`<=>(x-3-5+2x)(x-3+5-2x)=0`
`<=>(3x-8)(2-x)=0`
`<=>` $\left[\begin{matrix} x=\dfrac{8}{3}\\ x=2\end{matrix}\right.$
Vậy `S={8/3;2}`
`7)(3x-5)^2-(x-4)^2=0`
`<=>(3x-5-x+4)(3x-5+x-4)=0`
`<=>(2x-1)(4x-9)=0`
`<=>[(2x-1=0),(4x-9=0):}`
`<=>[(x=1/2),(x=9/4):}`
`8)(x^2-6x-9)-(5-2x)^2=0`
`<=>(x-3)^2-(5-2x)^2=0`
`<=>(x-3-5+2x)(x-3+5-2x)=0`
`<=>(3x-8)(2-x)=0`
`<=>[(3x-8=0),(2-x=0):}`
`<=>[(x=8/3),(x=2):}`
Vậy `S={2,8/3}`
