5. Suppose that $a \in \mathbf{R}$, line $(1-a) x+$ $(a+1) y-4(a+1)=0$, always passes through a fixed point $P$, and point $Q$ is on the curve $x^{2}-$ $x y+1=0$. Then the range of slope of a line passing through $P$ and $Q$ is ( ).
(A) $[-2,+\infty)$
(B) $[-3,+\infty)$
(C) $(1,+\infty)$
(D) $(3,+\infty)$
Note: fixed point - a fixed point, range - range, slope - slope, to pass through - to pass through
5.B.

The problem states: For any real number $a$, the line $(1-a) x + (a+1) y - 4(a+1) = 0$ always passes through a fixed point $P$. Point $Q$ lies on the curve $x^2 - xy + 1 = 0$. The range of the slope of the line passing through points $P$ and $Q$ is ( ).
(A) $[-2, +\infty)$
(B) $[-3, +\infty)$
(C) $(1, +\infty)$
(D) $(3, +\infty)$
Since the line passes through the fixed point $P$, we have
$(-x + y - 4) a + x + y - 4 = 0$.
Thus, $\left\{\begin{array}{l}-x + y - 4 = 0, \\ x + y - 4 = 0 .\end{array}\right.$ Solving this, we get $\left\{\begin{array}{l}x = 0, \\ y = 4 .\end{array}\right.$
Hence, $P(0, 4)$.
Let $Q(x, y)$. Since point $Q$ lies on $x^2 - xy + 1 = 0$, we have $x \neq 0, y = \frac{x^2 + 1}{x}$.
Therefore, the slope of the line passing through points $P$ and $Q$ is
$$
\begin{array}{l}
\frac{y - 4}{x - 0} = \frac{x^2 - 4x + 1}{x^2} = \frac{1}{x^2} - \frac{4}{x} + 1 \\
= \left(\frac{1}{x} - 2\right)^2 - 3 \geqslant -3 .
\end{array}
$$