Sure, let's solve these equations step by step.

### 1) \(\frac{X+2}{x} - \frac{x^2 + 5x + 4}{x(x+2)} = \frac{x}{x+2}\)

First, let's simplify the left-hand side (LHS):

\[
\frac{X+2}{x} - \frac{x^2 + 5x + 4}{x(x+2)}
\]

Notice that \(x^2 + 5x + 4\) can be factored as \((x+1)(x+4)\):

\[
\frac{X+2}{x} - \frac{(x+1)(x+4)}{x(x+2)}
\]

Now, let's find a common denominator for the terms on the LHS:

\[
\frac{X+2}{x} - \frac{(x+1)(x+4)}{x(x+2)} = \frac{(x+2)(x+2) - (x+1)(x+4)}{x(x+2)}
\]

Simplify the numerator:

\[
(x+2)(x+2) - (x+1)(x+4) = x^2 + 4x + 4 - (x^2 + 5x + 4) = x^2 + 4x + 4 - x^2 - 5x - 4 = -x
\]

So, the LHS simplifies to:

\[
\frac{-x}{x(x+2)} = \frac{-1}{x+2}
\]

Now, let's compare this with the right-hand side (RHS):

\[
\frac{-1}{x+2} = \frac{x}{x+2}
\]

For these to be equal, \(x\) must be \(-1\):

\[
\frac{-1}{x+2} = \frac{-1}{x+2}
\]

So, the solution is \(x = -1\).

### 2) \(\frac{1}{x+1} - \frac{5}{x+2} = \frac{15}{(x+1)(2-x)}\)

First, let's find a common denominator for the LHS:

\[
\frac{1}{x+1} - \frac{5}{x+2} = \frac{(x+2) - 5(x+1)}{(x+1)(x+2)}
\]

Simplify the numerator:

\[
(x+2) - 5(x+1) = x + 2 - 5x - 5 = -4x - 3
\]

So, the LHS becomes:

\[
\frac{-4x - 3}{(x+1)(x+2)}
\]

Now, let's compare this with the RHS:

\[
\frac{-4x - 3}{(x+1)(x+2)} = \frac{15}{(x+1)(2-x)}
\]

Notice that \((2-x)\) can be written as \(-(x-2)\), so:

\[
\frac{15}{(x+1)(2-x)} = \frac{15}{-(x+1)(x-2)} = -\frac{15}{(x+1)(x-2)}
\]

So, we have:

\[
\frac{-4x - 3}{(x+1)(x+2)} = -\frac{15}{(x+1)(x-2)}
\]

Cross-multiplying gives:

\[
-4x - 3 = -15(x+2)
\]

Simplify:

\[
-4x - 3 = -15x - 30
\]

Rearrange terms:

\[
-4x + 15x = -30 + 3
\]

\[
11x = -27
\]

\[
x = -\frac{27}{11}
\]

### 3) \(1 - \frac{6x}{x-2} + \frac{9x + 4}{x+2} = \frac{x(3x-2) + 1}{x^2 - 4}\)

First, let's simplify the LHS:

\[
1 - \frac{6x}{x-2} + \frac{9x + 4}{x+2}
\]

Notice that \(x^2 - 4\) can be factored as \((x-2)(x+2)\), so let's find a common denominator:

\[
1 - \frac{6x}{x-2} + \frac{9x + 4}{x+2} = \frac{(x-2)(x+2) - 6x(x+2) + (9x+4)(x-2)}{(x-2)(x+2)}
\]

Simplify the numerator:

\[
(x-2)(x+2) = x^2 - 4
\]

\[
-6x(x+2) = -6x^2 - 12x
\]

\[
(9x+4)(x-2) = 9x^2 - 18x + 4x - 8 = 9x^2 - 14x - 8
\]

Combine these:

\[
x^2 - 4 - 6x^2 - 12x + 9x^2 - 14x - 8 = 4x^2 - 26x - 12
\]

So, the LHS becomes:

\[
\frac{4x^2 - 26x - 12}{x^2 - 4}
\]

Now, let's compare this with the RHS:

\[
\frac{x(3x-2) + 1}{x^2 - 4}
\]

Simplify the numerator of the RHS:

\[
x(3x-2) + 1 = 3x^2 - 2x + 1
\]

So, we have:

\[
\frac{4x^2 - 26x - 12}{x^2 - 4} = \frac{3x^2 - 2x + 1}{x^2 - 4}
\]

For these to be equal, the numerators must be equal:

\[
4x^2 - 26x - 12 = 3x^2 - 2x + 1
\]

Rearrange terms:

\[
4x^2 - 3x^2 - 26x + 2x - 12 - 1 = 0
\]

\[
x^2 - 24x - 13 = 0
\]

Solve this quadratic equation using the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\):

\[
x = \frac{24 \pm \sqrt{24^2 - 4 \cdot 1 \cdot (-13)}}{2 \cdot 1}
\]

\[
x = \frac{24 \pm \sqrt{576 + 52}}{2}
\]

\[
x = \frac{24 \pm \sqrt{628}}{2}
\]

\[
x = \frac{24 \pm 2\sqrt{157}}{2}
\]

\[
x = 12 \pm \sqrt{157}
\]

So, the solutions are \(x = 12 + \sqrt{157}\) and \(x = 12 - \sqrt{157}\).