Break the absolute value inequality into two separate inequalities.

If $|u| > a$, then $u > a$ or $u < -a$.
So, we have:

The solution is $x > 1$ or $x < -2$.

The correct option is A.
$\boxed{\text{A. } x > 1 \text{ or } x < -2}$

20. If $A \subseteq B$ and $B \subseteq A$, then:

• $A \subseteq B$ means that every element in set $A$ is also an element in set $B$.
• $B \subseteq A$ means that every element in set $B$ is also an element in set $A$.

If $A$ is a subset of $B$ and $B$ is a subset of $A$, it implies that both sets contain exactly the same elements. This is the definition of set equality.

Therefore, $A = B$.

The correct option is B.
$\boxed{\text{B. } A = B}$

21. Rationalize: $\frac{1}{\sqrt{5} - 2}$

The denominator is $\sqrt{5} - 2$. Its conjugate is $\sqrt{5} + 2$.

In the numerator: $1 \times (\sqrt{5} + 2) = \sqrt{5} + 2$.
In the denominator, use the difference of squares formula $(a - b)(a + b) = a^2 - b^2$:

The correct option is A.
$\boxed{\text{A. } \sqrt{5} + 2}$

22. If $z = 2 - 2i$, find $|z|$

For a complex number $z = a + bi$, $a$ is the real part and $b$ is the imaginary part.
Here, $z = 2 - 2i$, so $a = 2$ and $b = -2$.

The modulus (or absolute value) of a complex number $z = a + bi$ is given by:

The modulus of $z$ is $\boxed{2\sqrt{2}}$.