The problem asks us to add and simplify the expression: $\frac{-3x}{x^2 - 4} + \frac{1}{x^2 - 4x + 4}$.

1. Problem Description

The problem asks us to add and simplify the expression: $\frac{-3x}{x^2 - 4} + \frac{1}{x^2 - 4x + 4}$.

2. Solution Steps

First, we factor the denominators:

$x^2 - 4 = (x-2)(x+2)$

$x^2 - 4x + 4 = (x-2)(x-2) = (x-2)^2$

Now, we rewrite the expression:

$\frac{-3x}{(x-2)(x+2)} + \frac{1}{(x-2)^2}$

To add these fractions, we need a common denominator. The least common denominator is $(x-2)^2(x+2)$.

We multiply the first fraction by $\frac{x-2}{x-2}$ and the second fraction by $\frac{x+2}{x+2}$:

$\frac{-3x(x-2)}{(x-2)^2(x+2)} + \frac{1(x+2)}{(x-2)^2(x+2)}$

$\frac{-3x^2 + 6x}{(x-2)^2(x+2)} + \frac{x+2}{(x-2)^2(x+2)}$

$\frac{-3x^2 + 6x + x + 2}{(x-2)^2(x+2)}$

$\frac{-3x^2 + 7x + 2}{(x-2)^2(x+2)}$

We try to factor the numerator. $-3x^2+7x+2$ doesn't seem to have $(x-2)$ or $(x+2)$ as a factor.

So, the final answer is:

$\frac{-3x^2 + 7x + 2}{(x-2)^2(x+2)}$

3. Final Answer

$\frac{-3x^2 + 7x + 2}{(x-2)^2(x+2)}$