d) \( \frac{x}{x-3}+\frac{6}{x^{2}-3 x}=1 \)

To solve the equation \( \frac{x}{x-3}+\frac{6}{x^{2}-3x}=1 \), we first notice that \( x^2 - 3x \) can be factored as \( x(x-3) \). Hence, we can rewrite the second term: \[ \frac{6}{x^2 - 3x} = \frac{6}{x(x-3)} \] Now, the equation becomes: \[ \frac{x}{x-3} + \frac{6}{x(x-3)} = 1 \] Next, we find a common denominator for these fractions, which is \( x(x-3) \): \[ \frac{x^2}{x(x-3)} + \frac{6}{x(x-3)} = 1 \] Combining the fractions on the left side gives: \[ \frac{x^2 + 6}{x(x-3)} = 1 \] To eliminate the fraction, multiply both sides by \( x(x-3) \) (assuming \( x \neq 0 \) and \( x \neq 3 \)): \[ x^2 + 6 = x(x-3) \] Expanding the right side: \[ x^2 + 6 = x^2 - 3x \] Now, we simplify by subtracting \( x^2 \) from both sides: \[ 6 = -3x \] Dividing both sides by -3 gives: \[ x = -2 \] Now, let's check if \( x = -2 \) satisfies the original equation. Substituting \( x = -2 \) back into the left side: \[ \frac{-2}{-2-3} + \frac{6}{(-2)^2 - 3(-2)} = \frac{-2}{-5} + \frac{6}{4 + 6} = \frac{2}{5} + \frac{6}{10} = \frac{2}{5} + \frac{3}{5} = 1 \] Since both sides are equal, the solution is confirmed. Thus, the final answer is: \[ \boxed{-2} \]