Đáp án:
`S=\{0\}`
Giải thích các bước giải:
`ĐKXĐ:x\ne 3;x\ne -3`
`(x-1)/(x+3)-(2x)/(x-3)=(7x-3)/(9-x^2)`
`⇔(x-1)/(x+3)-(2x)/(x-3)=(-(7x-3))/(x^2-9)`
`⇔((x-1)(x-3))/((x+3)(x-3))-(2x(x+3))/((x-3)(x+3))=(3-7x)/(x^2-9)`
`⇒(x-1)(x-3)-2x(x+3)=3-7x`
`⇔x^2-3x-x+3-2x^2-6x=3-7x`
`⇔x^2-4x+3-2x^2-6x=3-7x`
`⇔-x^2-10x+3=3-7x`
`⇔-x^2-10x+7x=3-3`
`⇔-x^2-3x=0`
`⇔-x(x+3)=0`
\(⇔\left[ \begin{array}{l}-x=0\\x+3=0\end{array} \right.\)
\(⇔\left[ \begin{array}{l}x=0(tm)\\x=-3(ktm)\end{array} \right.\)
Vậy `S=\{0\}`