\(\frac{x-3}{x+3}-\frac{x}{x-3}=\frac{18}{x^2-9}\)
ĐKXĐ : x \(\ne\) 3 ; x \(\ne\) - 3
\(\frac{x-3}{x+3}-\frac{x}{x-3}=\frac{18}{x^2-9}\)
<=> \(\frac{\left(x-3\right)^2}{x^2-9}-\frac{x\left(x+3\right)}{x^2-9}=\frac{18}{x^2-9}\)
Suy ra: (x - 3)2 - x(x + 3) = 18
<=.> x2 - 6x + 9 - x2 - 3x = 18
<=> x2 - 6x - x2 - 3x = 18 - 9
<=> -9x = 9
<=> x = -1 (nhận)
Vậy S = {-1}