lalo82 lalo82
  • 04-08-2018
  • Mathematics
contestada

(1/x-2)-(1/x+2)=
(A)=0
(B)=-(1/4)
(C)=(-4/x^(2))
(D)=(4/x^(2)-4)
(E)=(2x/x^(2)-4)
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Respuesta :

gmany
gmany gmany
  • 04-08-2018

[tex] \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) [/tex]

Used: [tex]a^2-b^2=(a-b)(a+b)[/tex]

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

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