To determine the value of \( k \) for which the lines \( L_1: x - (k+1)y + 2 = 0 \) and \( L_2: 2x - (k-1)y + 1 = 0 \) are parallel, we need to ensure that their slopes are equal. The slope of a line in the form \( ax + by + c = 0 \) is given by \( -\frac{a}{b} \). Therefore, we will set the slopes of both lines equal to each other and solve for \( k \).
Para que las rectas \( L_1: x - (k+1)y + 2 = 0 \) y \( L_2: 2x - (k-1)y + 1 = 0 \) sean paralelas, sus pendientes deben ser iguales. La pendiente de \( L_1 \) se calcula como:
\[
m_1 = -\frac{1}{k + 1}
\]
Y la pendiente de \( L_2 \) es:
\[
m_2 = -\frac{2}{k - 1}
\]
Igualamos las pendientes:
\[
-\frac{1}{k + 1} = -\frac{2}{k - 1}
\]
Eliminamos el signo negativo y multiplicamos ambos lados por \( (k + 1)(k - 1) \) para despejar la fracción:
\[
(k - 1) = 2(k + 1)
\]
Desarrollamos la ecuación:
\[
k - 1 = 2k + 2
\]
Reorganizamos para encontrar \( k \):
\[
k - 2k = 2 + 1
\]
\[
-k = 3
\]
\[
k = -3
\]
La solución para \( k \) es \\(\boxed{k = -3}\\).