Let's solve the three rational equations step by step.

Problem 1:

$\frac{x}{x+3} = 12$

Solution:

1. Multiply both sides by $x + 3$ to eliminate the denominator: $x = 12(x + 3)$
2. Expand the right-hand side: $x = 12x + 36$
3. Bring all terms involving $x$ to one side: $x - 12x = 36$
4. Simplify: $-11x = 36$
5. Divide by $-11$: $x = \frac{36}{-11} = -\frac{36}{11}$

Problem 2:

$\frac{x - 2}{x + 2} = \frac{x}{6}$

Solution:

1. Cross-multiply to eliminate the fractions: $6(x - 2) = x(x + 2)$
2. Expand both sides: $6x - 12 = x^2 + 2x$
3. Bring all terms to one side to set the equation to zero: $x^2 + 2x - 6x + 12 = 0$ $x^2 - 4x + 12 = 0$
4. Solve this quadratic equation using the quadratic formula: $x = \frac{-(-4) \pm \sqrt{(-4)^2 - 4(1)(12)}}{2(1)}$ $x = \frac{4 \pm \sqrt{16 - 48}}{2}$ $x = \frac{4 \pm \sqrt{-32}}{2}$ $x = \frac{4 \pm 4i\sqrt{2}}{2}$ $x = 2 \pm 2i\sqrt{2}$

Thus, the solutions are complex numbers: $x = 2 + 2i\sqrt{2}, \quad x = 2 - 2i\sqrt{2}$

Problem 3:

$\frac{x}{x+2} = \frac{8}{x^2 - 4}$

Solution:

1. Notice that $x^2 - 4$ is a difference of squares, so: $x^2 - 4 = (x - 2)(x + 2)$ The equation becomes: $\frac{x}{x+2} = \frac{8}{(x - 2)(x + 2)}$
2. Multiply both sides by $(x - 2)(x + 2)$ to eliminate the denominators: $x(x - 2) = 8$
3. Expand the left-hand side: $x^2 - 2x = 8$
4. Bring all terms to one side: $x^2 - 2x - 8 = 0$
5. Solve the quadratic equation using the quadratic formula: $x = \frac{-(-2) \pm \sqrt{(-2)^2 - 4(1)(-8)}}{2(1)}$ $x = \frac{2 \pm \sqrt{4 + 32}}{2}$ $x = \frac{2 \pm \sqrt{36}}{2}$ $x = \frac{2 \pm 6}{2}$
6. Thus, the solutions are: $x = \frac{2 + 6}{2} = 4, \quad x = \frac{2 - 6}{2} = -2$

However, $x = -2$ is not valid because it would make the original denominator zero. So the only solution is: $x = 4$

Let me know if you'd like any more details! Here are some related questions:

1. What are the conditions for the existence of a solution in rational equations?
2. How do you handle complex numbers when solving quadratic equations?
3. Can a rational equation have no solution? If so, why?
4. How does cross-multiplying work in rational equations?
5. Why must denominators in rational equations not equal zero?

Tip: Always check for extraneous solutions, especially when dealing with denominators, as some solutions might make the denominator zero and invalidate them.