Respostas

2014-06-29T13:53:41-03:00
 \frac{1}{x-2} + \frac{1}{x+2} = \frac{3- x^{2} }{ x^{2} -4}  \\ \\   \frac{(x+2)+(x-2)}{(x-2)(x+2)} = \frac{3- x^{2} }{ x^{2} -4} \\  \\  \frac{2x}{ x^{2} -4} =\frac{3- x^{2} }{ x^{2} -4} \\  \\ 2x=3- x^{2}  \\ 2x+ x^{2} -3=0 \\  x^{2} +2x-3=0

Δ = b² - 4.a.c 
Δ = 2² - 4 . 1 . -3 
Δ = 4 - 4. 1 . -3 
Δ = 16

x = (-b +- √Δ)/2a
x' = (-2 + √16)/2.1       x'' = (-2 - √16)/2.1
x' = 2 / 2                     x'' = -6 / 2
x' = 1                          x'' = -3