Đáp án: `N=\frac{x-3}{x+3}`
Giải thích các bước giải:
$ĐKXĐ:x\neq±3;0$
`N=(\frac{3}{x^2+6x+9}-\frac{1}{x+3})÷(\frac{3}{x^2-9}-\frac{1}{x-3})`
`=[\frac{3}{(x+3)^2}-\frac{1}{x+3}]÷[\frac{3}{(x-3)(x+3)}-\frac{1}{x-3}]`
`=\frac{3-(x+3)}{(x+3)^2}÷\frac{3-(x+3)}{(x-3)(x+3)}`
`=\frac{x}{(x+3)^2}.\frac{(x-3)(x+3)}{x}=\frac{x-3}{x+3}`