To solve the equation

\[ \frac{2}{x^2 + 3x + 2} + \frac{1}{x^2 + 5x + 6} = \frac{1}{x^2 - 3x + 2}, \]

we first factor the quadratic expressions in the denominators.

1. Factor \(x^2 + 3x + 2\):
\[ x^2 + 3x + 2 = (x + 1)(x + 2). \]

2. Factor \(x^2 + 5x + 6\):
\[ x^2 + 5x + 6 = (x + 2)(x + 3). \]

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

Substitute these factorizations back into the equation:

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

To combine the fractions on the left-hand side, we need a common denominator, which is \((x + 1)(x + 2)(x + 3)\):

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

Simplify the numerator on the left-hand side:

\[ \frac{2(x + 3) + (x + 1)}{(x + 1)(x + 2)(x + 3)} = \frac{2x + 6 + x + 1}{(x + 1)(x + 2)(x + 3)} = \frac{3x + 7}{(x + 1)(x + 2)(x + 3)}. \]

So the equation becomes:

\[ \frac{3x + 7}{(x + 1)(x + 2)(x + 3)} = \frac{1}{(x - 1)(x - 2)}. \]

Cross-multiply to solve for \(x\):

\[ (3x + 7)(x - 1)(x - 2) = (x + 1)(x + 2)(x + 3). \]

Expand both sides:

Left-hand side:
\[ (3x + 7)(x^2 - 3x + 2) = 3x^3 - 9x^2 + 6x + 7x^2 - 21x + 14 = 3x^3 - 2x^2 - 15x + 14. \]

Right-hand side:
\[ (x + 1)(x + 2)(x + 3) = (x + 1)(x^2 + 5x + 6) = x^3 + 5x^2 + 6x + x^2 + 5x + 6 = x^3 + 6x^2 + 11x + 6. \]

Set the expanded forms equal to each other:

\[ 3x^3 - 2x^2 - 15x + 14 = x^3 + 6x^2 + 11x + 6. \]

Combine like terms:

\[ 3x^3 - 2x^2 - 15x + 14 - x^3 - 6x^2 - 11x - 6 = 0, \]

\[ 2x^3 - 8x^2 - 26x + 8 = 0. \]

This is a cubic equation. Solving cubic equations analytically can be complex, so we might use numerical methods or graphing to find the roots. However, we can check for rational roots using the Rational Root Theorem, which suggests that any rational root, \(p/q\), is a factor of the constant term (8) divided by a factor of the leading coefficient (2).

Possible rational roots are \( \pm 1, \pm 2, \pm 4, \pm 8, \pm \frac{1}{2}, \pm \frac{4}{2} \).

Testing these values, we find:

For \(x = 1\):
\[ 2(1)^3 - 8(1)^2 - 26(1) + 8 = 2 - 8 - 26 + 8 = -24 \neq 0. \]

For \(x = -1\):
\[ 2(-1)^3 - 8(-1)^2 - 26(-1) + 8 = -2 - 8 + 26 + 8 = 24 \neq 0. \]

For \(x = 2\):
\[ 2(2)^3 - 8(2)^2 - 26(2) + 8 = 16 - 32 - 52 + 8 = -60 \neq 0. \]

For \(x = -2\):
\[ 2(-2)^3 - 8(-2)^2 - 26(-2) + 8 = -16 - 32 + 52 + 8 = 12 \neq 0. \]

For \(x = 4\):
\[ 2(4)^3 - 8(4)^2 - 26(4) + 8 = 128 - 128 - 104 + 8 = -96 \neq 0. \]

For \(x = -4\):
\[ 2(-4)^3 - 8(-4)^2 - 26(-4) + 8 = -128 - 128 + 104 + 8 = -144 \neq 0. \]

For \(x = 8\):
\[ 2(8)^3 - 8(8)^2 - 26(8) + 8 = 1024 - 512 - 208 + 8 = 312 \neq 0. \]

For \(x = -8\):
\[ 2(-8)^3 - 8(-8)^2 - 26(-8) + 8 = -1024 - 512 + 208 + 8 = -1320 \neq 0. \]

For \(x = \frac{1}{2}\):
\[ 2\left(\frac{1}{2}\right)^3 - 8\left(\frac{1}{2}\right)^2 - 26\left(\frac{1}{2}\right) + 8 = \frac{2}{8} - \frac{8}{4} - 13 + 8 = \frac{1}{4} - 2 - 13 + 8 = -6.75 \neq 0. \]

For \(x = -\frac{1}{2}\):
\[ 2\left(-\frac{1}{2}\right)^3 - 8\left(-\frac{1}{2}\right)^2 - 26\left(-\frac{1}{2}\right) + 8 = -\frac{2}{8} - \frac{8}{4} + 13 + 8 = -\frac{1}{4} - 2 + 13 + 8 = 18.75 \neq 0. \]

For \(x = \frac{4}{2} = 2\):
\[ 2(2)^3 - 8(2)^2 - 26(2) + 8 = 16 - 32 - 52 + 8 = -60 \neq 0. \]

None of the rational roots work, so we may need to use numerical methods or graphing to find the roots.