Five, (Full marks 12 points) Find all values of $k$ such that the roots of the equation $k x^{2}+(k+1) x+(k-1)=0$ are integers.

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When $k=0$, the given equation is a linear equation $x-1=0$, which has an integer root 1.
When $k \neq 0$, the given equation is a quadratic equation.
Let the two integer roots be $x_{1}$ and $x_{2}$, then we have
$$
\left\{\begin{array}{l}
x_{1}+x_{2}=-\frac{k+1}{k}=-1-\frac{1}{k}, \\
x_{1} x_{2}=\frac{k-1}{b}=1-\frac{1}{k}.
\end{array}\right.
$$

From (1)-(2) we get $x_{1}+x_{2}-x_{1} x_{2}=-2$.
Thus, $\left(x_{1}-1\right)\left(x_{2}-1\right)=3=1 \times 3$,
$$
\begin{array}{c}
\therefore\left\{\begin{array}{l}
x_{1}-1=1, \\
x_{2}-1=3;
\end{array} \quad \left\{\begin{array}{l}
x_{1}-1=-1, \\
x_{2}-1=-3;
\end{array}\right.\right. \\
\left\{\begin{array}{l}
x_{1}-1=3, \\
x_{2}-1=1;
\end{array} \left\{\begin{array}{l}
x_{1}-1=-3, \\
x_{2}-1=-1.
\end{array}\right.\right. \\
\therefore x_{1}+x_{2}=6 \text { or } x_{1}+x_{2}=-2. \\
\therefore-1-\frac{1}{k}=6 \text { or }-1-\frac{1}{k}=-2.
\end{array}
$$

Solving, we get $k=-\frac{1}{7}$ or $k=1$.
$$
\text { Also, } \Delta=(k+1)^{2}-4 k(k-1)=-3 k^{2}+6 k+1 \text {, }
$$

When $k=-\frac{1}{7}$ and $k=1$, $\Delta>0$ in both cases.
Thus, the values of $k$ that satisfy the requirements are $k=0, k=-\frac{1}{7}, k=1$.