Answer
$\dfrac{x+2}{x-2}, \space x\ne-2$
Work Step by Step
Note that $\dfrac{x^2+4x+4}{x^2-4}=\dfrac{x^2+2(x)(2)+2^2}{x^2-2^2}$
Use special formula $a^2+2ab+b^2=(a+b)^2$ in the numerator and $a^2-b^2=(a+b)(a-b)$ in the denominator to obtain:.
$=\dfrac{(x+2)^2}{(x+2)(x-2)}$
$=\dfrac{(x+2)(x+2)}{(x+2)(x-2)}$
Cancel the common factor $x+2$:
$=\dfrac{x+2}{x-2}, x\ne-2$
Hence, the lowest term is $\dfrac{x+2}{x-2}, x\ne-2$.
