Let's simplify and solve the two equations given.

1. **First equation:**
\[
X(x + 1) = 2(x^2 - 1)b
\]

We can expand this equation:
\[
Xx + X = 2b(x^2 - 1)
\]
\[
Xx + X = 2bx^2 - 2b
\]
Rearranging gives us:
\[
2bx^2 - Xx - (X + 2b) = 0
\]

This is a quadratic equation in the form \( ax^2 + bx + c = 0 \) where:
- \( a = 2b \)
- \( b = -X \)
- \( c = -(X + 2b) \)

2. **Second equation:**
\[
x(x + 1) = 2(x \mu 2 - 1)
\]

Assuming "x mu 2" means multiplication:
\[
x(x + 1) = 2(2x - 1)
\]
Expanding this yields:
\[
x^2 + x = 4x - 2
\]
Rearranging gives us:
\[
x^2 + x - 4x + 2 = 0
\]
This simplifies to:
\[
x^2 - 3x + 2 = 0
\]

This factors to:
\[
(x - 1)(x - 2) = 0
\]

Therefore, the solutions for \( x \) are:
\[
x = 1 \quad \text{or} \quad x = 2
\]

If you need further assistance or a detailed explanation about any specific part, please let me know!