Problem 22

To solve the equation $\frac{{x}^{2}-4}{x+2}=0$, we start by setting the numerator equal to zero, as the value of a fraction is zero if and only if its numerator is zero. Thus, we have:

$$x^{2}-4=0$$

This can be factored into:

$$(x-2)(x+2)=0$$

From this, we find two possible values for $x$:

1. $x-2=0 \Rightarrow x=2$
2. $x+2=0 \Rightarrow x=-2$

However, we must also consider the denominator of the original fraction, $x+2$, to ensure it is not zero, as division by zero is undefined.

- When $x=2$, the denominator $x+2=4$, which is not equal to zero. Therefore, $x=2$ is a valid solution.
- When $x=-2$, the denominator $x+2=0$, which would make the fraction undefined. Therefore, $x=-2$ is not a valid solution.

Thus, the only value of $x$ for which the fraction $\frac{{x}^{2}-4}{x+2}$ is $0$ is $x=2$.

$$\boxed{x=2}$$