(1/x-2)-(1/x+2)=

1 answer:

5 0

$\dfrac{1}{x-2}-\dfrac{1}{x+2}=\dfrac{x+2}{(x-2)(x+2)}-\dfrac{x-2}{(x-2)(x+2)}\\\\=\dfrac{(x+2)-(x-2)}{(x-2)(x+2)}=\dfrac{x+2-x+2}{(x-2)(x+2)}=\dfrac{4}{x^2-4}\to(D)$

Used: $a^2-b^2=(a-b)(a+b)$

Answer: (D)=(4/x^(2)-4)

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