The problem asks us to prove that the solution of the equation $\frac{3m}{9m^2 - 12mx + 4x^2} - \frac{4x - 3m}{9m^2 - 4x^2} = \frac{2}{2x + 3m}$ does not depend on the real parameter $m$, given that $m \ne 0$ and $x \ne \pm \frac{3m}{2}$.

1. Problem Description

The problem asks us to prove that the solution of the equation

$\frac{3m}{9m^2 - 12mx + 4x^2} - \frac{4x - 3m}{9m^2 - 4x^2} = \frac{2}{2x + 3m}$

does not depend on the real parameter $m$, given that $m \ne 0$ and $x \ne \pm \frac{3m}{2}$.

2. Solution Steps

First, we factor the denominators:

$9m^2 - 12mx + 4x^2 = (3m - 2x)^2 = (2x - 3m)^2$

$9m^2 - 4x^2 = (3m - 2x)(3m + 2x)$

Now we can rewrite the equation:

$\frac{3m}{(2x - 3m)^2} - \frac{4x - 3m}{(3m - 2x)(3m + 2x)} = \frac{2}{2x + 3m}$

$\frac{3m}{(2x - 3m)^2} + \frac{4x - 3m}{(2x - 3m)(2x + 3m)} = \frac{2}{2x + 3m}$

Multiply both sides by $(2x - 3m)^2(2x + 3m)$:

$3m(2x + 3m) + (4x - 3m)(2x - 3m) = 2(2x - 3m)^2$

$6mx + 9m^2 + 8x^2 - 12mx - 6mx + 9m^2 = 2(4x^2 - 12mx + 9m^2)$

$8x^2 - 12mx + 18m^2 = 8x^2 - 24mx + 18m^2$

$8x^2 - 12mx + 18m^2 - (8x^2 - 24mx + 18m^2) = 0$

$12mx = 0$

Since $m \ne 0$, we have $x = 0$.

However, the problem states that $x \ne \pm \frac{3m}{2}$. Let's check if $x=0$ satisfies this condition. If $x=0$, then $0 \ne \pm \frac{3m}{2}$. This is true as long as $m\ne0$. Thus, the solution $x=0$ is valid.

Since the solution $x = 0$ does not depend on $m$, we have shown that the solution of the equation does not depend on the real parameter $m$.

3. Final Answer

The solution is $x = 0$. Since this solution does not depend on $m$, the solution of the equation does not depend on the real parameter $m$.