# lowest terms app1/x^2-4+ 1/x^2-4x+4 -4/(x+2)(x-2)^2

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### 1 Answer

You should notice that each denominator is a special product such that:

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

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

Writing the fractions again yields:

`1/((x-2)(x+2)) + 1/((x-2)^2) - 4/((x+2)(x-2)^2)`

You need to bring the fractions to the common denominator such that:

`(x-2)/((x+2)(x-2)^2) + (x+2)/((x+2)(x-2)^2) - 4/((x+2)(x-2)^2)`

Adding the numerators yields:

`(x - 2 + x + 2 - 4)/((x+2)(x-2)^2) = (2x - 4)/((x+2)(x-2)^2)`

Factoring out 2 at numerator yields:

`2(x - 2)/((x+2)(x-2)^2)`

Reducing by x - 2 yields:

`2/((x+2)(x-2)) = 2/(x^2 - 4)`

**Hence, reducing the fractions to the lowest terms yields `2/(x^2 - 4).` **