39. If the solution of the equation $\frac{2 x-a}{x-1}+\frac{3}{1-x}=4$ with respect to $\mathrm{x}$ is non-negative, then the range of values for $a$ is
Answer: $a \leq 1$ and $a \neq -1$
Solution: $\frac{2 x-a}{x-1}+\frac{3}{1-x}=4$ can be transformed into
$$
\frac{2 x-a-3}{x-1}=4,
$$

Eliminating the denominator, we get
$$
2 x-a-3=4 x-4 \text {, }
$$

Solving for $x$, we get
$$
x=\frac{1-a}{2} \text {, }
$$

Since the solution of the equation is non-negative,
we have
$$
\frac{1-a}{2} \geq 0 \text { and } \frac{1-a}{2} \neq 1,
$$

Solving for $a$, we get
$a \leq 1$ and $a \neq -1$.