`d)` $\text{ĐKXĐ:} \begin{cases}x-3\ne 0\\ x+3\ne 0\\ x^2-9\ne 0\\\end{cases} ⇔ \begin{cases} x-3\ne 0\\ x+3 \ne 0\\ (x-3)(x+3)\ne 0\\\end{cases} ⇔ x \ne \pm 3.$
`\frac{x-1}{x-3}+\frac{x+2}{x+3}=\frac{4x-9}{x^2-9}`
`⇔ \frac{x-1}{x-3}+\frac{x+2}{x+3}-\frac{4x-9}{x^2-9}=0`
`⇔ \frac{(x-1)(x+3)+(x+2)(x-3)-(4x-9)}{(x-3)(x+3)}=0`
`⇔ x^2+2x-3+x^2-x-6-4x+9=0`
`⇔ 2x^2 -3x=0`
`⇔ x(2x-3)=0`
`⇔` \(\left[ \begin{array}{l}x=0\\2x-3=0\end{array} \right.\) `⇔` \(\left[ \begin{array}{l}x=0\\x=\frac{3}{2}\end{array} \right.\)
`\to` Vậy `S={\frac{3}{2};0}`
