a) `5x(x-7)-x+7=0`
`⇔5x(x-7)-(x-7)=0`
`⇔(x-7)(5x-1)=0`
`⇔`\(\left[ \begin{array}{l}x-7=0\\5x-1=0\end{array} \right.\)
`⇔`\(\left[ \begin{array}{l}x=7\\5x=1\end{array} \right.\)
`⇔`\(\left[ \begin{array}{l}x=7\\x=\dfrac{1}{5}\end{array} \right.\)
Vậy: `S={7;\frac{1}{5}}`
b) `3(x+3)-x^2-3x=0`
`⇔3(x+3)-x(x+3)=0`
`⇔(x+3)(3-x)=0`
`⇔`\(\left[ \begin{array}{l}x+3=0\\3-x=0\end{array} \right.\)
`⇔`\(\left[ \begin{array}{l}x=-3\\x=3\end{array} \right.\)
Vậy: `S={-3;3}`
c) `(x+3)^2-(x-4)(x+4)=0`
`⇔(x+3)^2-(x^2-16)=0`
`⇔x^2+6x+9-x^2+16=0`
`⇔6x+25=0`
`⇔6x=-25`
`⇔x=-\frac{25}{6}`
Vậy: `S={-\frac{25}{6}}`
