Solve the rational equation \( \frac{x}{x-3} - \frac{2}{x+1} = \frac{1}{x^2-2x-3} \).

Step 1 ：The first step is to simplify the equation. We can do this by factoring the denominator in the right side, \(x^2-2x-3\), which factors into \((x-3)(x+1)\), thus the equation becomes \( \frac{x}{x-3} - \frac{2}{x+1} = \frac{1}{(x-3)(x+1)} \).

Step 2 ：Next, we multiply every term by the common denominator, \((x-3)(x+1)\), of the fractions to get rid of the fractions. This gives us \(x(x+1) - 2(x-3) = 1\).

Step 3 ：Expanding and simplifying the equation gives us \(x^2 + x -2x + 6 = 1\), which simplifies to \(x^2 - x + 6 = 1\).

Step 4 ：Subtracting 1 from both sides gives us the quadratic equation \(x^2 - x + 5 = 0\).

Step 5 ：We can solve this quadratic equation using the quadratic formula, \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 1\), \(b = -1\), and \(c = 5\).

Step 6 ：Substituting these values into the formula gives us \(x = \frac{1 \pm \sqrt{(-1)^2 - 4(1)(5)}}{2(1)}\), which simplifies to \(x = \frac{1 \pm \sqrt{1 - 20}}{2}\), and further simplifies to \(x = \frac{1 \pm \sqrt{-19}}{2}\).

Step 7 ：Since the square root of a negative number is not a real number, there are no real solutions to the rational equation.