Let's solve the given equation step by step. The equation is:

\[
\frac{4x}{4x-3} + \frac{5}{3x-4x^2} = \frac{x+2}{x}
\]

First, let's simplify the terms on the left-hand side.

1. Simplify \(\frac{4x}{4x-3}\):
This term is already in its simplest form.

2. Simplify \(\frac{5}{3x-4x^2}\):
Factor out \(x\) from the denominator:
\[
3x - 4x^2 = x(3 - 4x)
\]
So, the term becomes:
\[
\frac{5}{x(3 - 4x)}
\]

Now, the equation looks like this:

\[
\frac{4x}{4x-3} + \frac{5}{x(3-4x)} = \frac{x+2}{x}
\]

Next, let's find a common denominator for the left-hand side. The common denominator for \(\frac{4x}{4x-3}\) and \(\frac{5}{x(3-4x)}\) is \((4x-3)x(3-4x)\).

Rewrite each fraction with the common denominator:

\[
\frac{4x \cdot x(3-4x)}{(4x-3)x(3-4x)} + \frac{5 \cdot (4x-3)}{(4x-3)x(3-4x)} = \frac{x+2}{x}
\]

Simplify the numerators:

\[
\frac{4x^2(3-4x) + 5(4x-3)}{(4x-3)x(3-4x)} = \frac{x+2}{x}
\]

Expand the numerators:

\[
\frac{4x^2(3-4x) + 5(4x-3)}{(4x-3)x(3-4x)} = \frac{x+2}{x}
\]

\[
\frac{12x^2 - 16x^3 + 20x - 15}{(4x-3)x(3-4x)} = \frac{x+2}{x}
\]

Combine the terms in the numerator:

\[
\frac{12x^2 - 16x^3 + 20x - 15}{(4x-3)x(3-4x)} = \frac{x+2}{x}
\]

Now, we can cross-multiply to solve for \(x\):

\[
(12x^2 - 16x^3 + 20x - 15)x = (x+2)(4x-3)x(3-4x)
\]

Since \(x \neq 0\), we can cancel \(x\) from both sides:

\[
12x^2 - 16x^3 + 20x - 15 = (x+2)(4x-3)(3-4x)
\]

Expand the right-hand side:

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

First, expand \((4x-3)(3-4x)\):

\[
(4x-3)(3-4x) = 12x - 16x^2 - 9 + 12x = -16x^2 + 24x - 9
\]

Now, multiply by \((x+2)\):

\[
(x+2)(-16x^2 + 24x - 9) = -16x^3 + 24x^2 - 9x - 32x^2 + 48x - 18 = -16x^3 - 8x^2 + 39x - 18
\]

So the equation becomes:

\[
12x^2 - 16x^3 + 20x - 15 = -16x^3 - 8x^2 + 39x - 18
\]

Combine like terms:

\[
12x^2 - 16x^3 + 20x - 15 + 16x^3 + 8x^2 - 39x + 18 = 0
\]

\[
20x^2 - 19x + 3 = 0
\]

Now, solve the quadratic equation \(20x^2 - 19x + 3 = 0\) using the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\):

Here, \(a = 20\), \(b = -19\), and \(c = 3\):

\[
x = \frac{-(-19) \pm \sqrt{(-19)^2 - 4 \cdot 20 \cdot 3}}{2 \cdot 20}
\]

\[
x = \frac{19 \pm \sqrt{361 - 240}}{40}
\]

\[
x = \frac{19 \pm \sqrt{121}}{40}
\]

\[
x = \frac{19 \pm 11}{40}
\]

So, the solutions are:

\[
x = \frac{19 + 11}{40} = \frac{30}{40} = \frac{3}{4}
\]

\[
x = \frac{19 - 11}{40} = \frac{8}{40} = \frac{1}{5}
\]

Therefore, the solutions to the equation are:

\[
x = \frac{3}{4} \quad \text{and} \quad x = \frac{1}{5}
\]