求满足 $\frac{x^2 + 2x + 2}{x^2 - 5x + 6} \ge 0.$ 的 $x$ 的取值范围。 证明对于所有实数 $k$，直线 $y = x + k$ 与曲线 $xy = y + 2$ 总是相交于两个不同的点。

1. Analyse the sign of the numerator

For the numerator $x^2 + 2x + 2$:
Discriminant $\Delta = 2^2 - 4(1)(2) = 4 - 8 = -4$.
Since $\Delta < 0$ and the coefficient of $x^2$ is $1 > 0$, $x^2 + 2x + 2 > 0$ for all real values of $x$.

2. Set up the inequality for the denominator

Since the numerator is always strictly positive, for the fraction $\frac{x^2 + 2x + 2}{x^2 - 5x + 6} \ge 0$ to hold, the denominator must be strictly positive (it cannot be zero).
$x^2 - 5x + 6 > 0$

3. Solve the quadratic inequality for the denominator

$(x - 2)(x - 3) > 0$
The critical values are $x = 2$ and $x = 3$.

4. Determine the final range of values for x

Since the quadratic curve opens upwards, the expression is positive outside the roots.
$x < 2 \text{ or } x > 3$

1. Substitute the equation of the line into the equation of the curve

Substitute $y = x + k$ into $xy = y + 2$:
$x(x + k) = (x + k) + 2$

2. Rearrange into a standard quadratic equation form $ax^2 + bx + c = 0$

$x^2 + kx = x + k + 2$
$x^2 + (k - 1)x - (k + 2) = 0$

3. Find the discriminant of the quadratic equation

$\Delta = b^2 - 4ac$
$\Delta = (k - 1)^2 - 4(1)[-(k + 2)]$
$\Delta = k^2 - 2k + 1 + 4k + 8$
$\Delta = k^2 + 2k + 9$

4. Show that the discriminant is always positive for all real $k$

Complete the square for the discriminant:
$\Delta = (k + 1)^2 - 1 + 9$
$\Delta = (k + 1)^2 + 8$
Since $(k + 1)^2 \ge 0$ for all real $k$, $(k + 1)^2 + 8 \ge 8 > 0$.
Therefore, $\Delta > 0$ for all real $k$, which means the equation always has two distinct real roots. Thus, the line always intersects the curve at two distinct points.