Cho 9x² + 4y² = 20xy, \( (3x - 2y > 0) \)

To solve the equation \(9x^2 + 4y^2 = 20xy\), we can rearrange it into a standard form suitable for analysis.

### Step 1: Rearranging the equation
We can rearrange the equation as follows:

\[
9x^2 - 20xy + 4y^2 = 0
\]

This is a quadratic equation in terms of \(x\).

### Step 2: Using the quadratic formula
For the quadratic \(ax^2 + bx + c = 0\), the quadratic formula gives us:

\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]

Here, \(a = 9\), \(b = -20y\), and \(c = 4y^2\).

### Step 3: Calculate the discriminant
Calculate the discriminant \(D\):

\[
D = b^2 - 4ac = (-20y)^2 - 4 \cdot 9 \cdot 4y^2 = 400y^2 - 144y^2 = 256y^2
\]

### Step 4: Finding the roots
Since the discriminant is positive, we have two real roots:

\[
x = \frac{20y \pm \sqrt{256y^2}}{2 \cdot 9} = \frac{20y \pm 16y}{18}
\]

This will yield two solutions:

1. \(x_1 = \frac{36y}{18} = 2y\)
2. \(x_2 = \frac{4y}{18} = \frac{2y}{9}\)

### Step 5: Consider the condition \(3x > 2y > 0\)
Given the condition \(3x > 2y\):

- For \(x_1 = 2y\):
\[
3(2y) > 2y \Rightarrow 6y > 2y \Rightarrow 4y > 0 \text{ (true if } y > 0)
\]

- For \(x_2 = \frac{2y}{9}\):
\[
3\left(\frac{2y}{9}\right) > 2y \Rightarrow \frac{6y}{9} > 2y \Rightarrow \frac{2y}{3} > 2y \Rightarrow \frac{2}{3} > 2 \text{ (not true)}
\]

Thus, the only valid solution is \(x = 2y\).

### Step 6: Proving \( \frac{6x - 4y}{3x + 2y} = 1\)
Now confirm that \( \frac{6x - 4y}{3x + 2y} = 1\) with \(x = 2y\):

Substituting \(x = 2y\):

\[
6x - 4y = 6(2y) - 4y = 12y - 4y = 8y
\]
\[
3x + 2y = 3(2y) + 2y = 6y + 2y = 8y
\]

Thus,

\[
\frac{6x - 4y}{3x + 2y} = \frac{8y}{8y} = 1
\]

### Conclusion
We have shown that if \(9x^2 + 4y^2 = 20xy\) and \(3x > 2y > 0\), then indeed \( \frac{6x - 4y}{3x + 2y} = 1\).