We are asked to solve the equation $\frac{2mx(m+1)+3x}{m^3-27} - \frac{x}{m-3} + \frac{x+1}{m^2+3m+9} = 0$ for $x$.

1. Problem Description

We are asked to solve the equation

$\frac{2mx(m+1)+3x}{m^3-27} - \frac{x}{m-3} + \frac{x+1}{m^2+3m+9} = 0$

for $x$.

2. Solution Steps

First, we factor the denominator $m^3 - 27$ using the difference of cubes formula:

$m^3 - 27 = m^3 - 3^3 = (m-3)(m^2+3m+9)$.

The equation can be rewritten as

$\frac{2mx(m+1)+3x}{(m-3)(m^2+3m+9)} - \frac{x}{m-3} + \frac{x+1}{m^2+3m+9} = 0$.

We multiply both sides of the equation by $(m-3)(m^2+3m+9)$ to eliminate the denominators:

$2mx(m+1)+3x - x(m^2+3m+9) + (x+1)(m-3) = 0$.

Expand the terms:

$2m^2x + 2mx + 3x - xm^2 - 3mx - 9x + xm - 3x + m - 3 = 0$.

Group the terms with $x$:

$(2m^2 - m^2 + 2m - 3m + m + 3 - 9 - 3)x + m - 3 = 0$

$(m^2 - 7)x + (m-3) = 0$.

Solve for $x$:

$(m^2 - 7)x = 3 - m$

$x = \frac{3-m}{m^2-7}$.

If $m^2 - 7 = 0$, then $m = \pm \sqrt{7}$.

If $m = \sqrt{7}$, then we have $0x = 3 - \sqrt{7}$ which has no solution.

If $m = -\sqrt{7}$, then we have $0x = 3 + \sqrt{7}$ which has no solution.

Therefore, we assume $m^2 - 7 \neq 0$.

3. Final Answer

$x = \frac{3-m}{m^2-7}$