Certainly! Let's solve each of the given equations step by step.

### a.

\[ \frac{3x}{2x+3} - \frac{6x}{4x-1} = 0 \]

To solve this equation, bring both fractions to a common denominator and equate the numerators:

1. Find a common denominator: \((2x+3)(4x-1)\).
2. Rewrite the equation:

\[
\frac{3x(4x-1) - 6x(2x+3)}{(2x+3)(4x-1)} = 0
\]

3. Solve the numerator:

\[
3x(4x-1) = 12x^2 - 3x \\
6x(2x+3) = 12x^2 + 18x
\]

\[
12x^2 - 3x - (12x^2 + 18x) = -3x - 18x = -21x
\]

So the equation is:

\[
\frac{-21x}{(2x+3)(4x-1)} = 0
\]

4. Equating the numerator to zero:

\[
-21x = 0 \implies x = 0
\]

### b.

\[ \frac{1}{2x-1} + \frac{3}{2x+1} = \frac{6x-3}{4x^2-1} \]

The right-hand side can be factored as:

\[ \frac{6x-3}{(2x-1)(2x+1)} \]

Bring the left-hand side to the same denominator:

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

\[
\frac{2x+1 + 6x - 3}{(2x-1)(2x+1)} = \frac{8x - 2}{(2x-1)(2x+1)}
\]

Since the denominators are the same, equate the numerators:

\[
8x - 2 = 6x - 3
\]

Solving gives:

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

### c.

\[ \frac{5x-1}{3x+2} - \frac{5x+2}{3x} = 0 \]

1. Bring both fractions to a common denominator: \( (3x+2)(3x) \).

2. Rewrite the equation:

\[
\frac{(5x-1)(3x) - (5x+2)(3x+2)}{3x(3x+2)} = 0
\]

3. Solve the numerator:

\[
(5x-1)(3x) = 15x^2 - 3x \\
(5x+2)(3x+2) = 15x^2 + 10x + 6x + 4 = 15x^2 + 16x + 4
\]

\[
15x^2 - 3x - (15x^2 + 16x + 4) = -3x - 16x - 4 = -19x - 4
\]

The equation now is:

\[
\frac{-19x - 4}{3x(3x+2)} = 0
\]

4. Equate the numerator to zero:

\[
-19x - 4 = 0 \implies x = -\frac{4}{19}
\]

### d.

\[ \frac{6x-5}{2x-1} - \frac{9x}{3x+1} = 0 \]

1. Bring both fractions to a common denominator: \( (2x-1)(3x+1) \).

2. Rewrite the equation:

\[
\frac{(6x-5)(3x+1) - 9x(2x-1)}{(2x-1)(3x+1)} = 0
\]

3. Solve the numerator:

\[
(6x-5)(3x+1) = 18x^2 + 6x - 15x - 5 = 18x^2 - 9x - 5
\]

\[
9x(2x-1) = 18x^2 - 9x
\]

\[
18x^2 - 9x - 5 - (18x^2 - 9x) = -5
\]

The numerator becomes \(-5\), which cannot be zero; hence, there's no solution in this form.

### e.

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

This is not set equal to anything, so it seems incomplete as an equation to solve for \(x\).

### g.

\[ \frac{3}{2x+1} + \frac{7}{3x+2} = \frac{21x+10}{(2x+1)(3x+2)} \]

Bring both fractions to a common denominator (already given):

\[
\frac{3(3x+2) + 7(2x+1)}{(2x+1)(3x+2)} = \frac{21x+10}{(2x+1)(3x+2)}
\]

Solve the numerators:

\[
3(3x+2) = 9x + 6 \\
7(2x+1) = 14x + 7
\]

\[
9x + 6 + 14x + 7 = 23x + 13
\]

Equating the numerators:

\[
23x + 13 = 21x + 10
\]

Solving gives:

\[
2x = -3 \implies x = -\frac{3}{2}
\]

These equations are now either solved or identified as problematic due to inconsistencies. If more equations exist, feel free to provide them.