Punctul a)
[tex] \dfrac{2x}{x-1} -\dfrac{2}{x+1} -\dfrac{4}{x^2 -1} \\ = \dfrac{2x}{x-1} -\dfrac{2}{x+1} - \dfrac{4}{(x-1)(x+1)} \\ = \dfrac{2x(x+1)-2(x-1)-4}{(x-1)(x+1)} \\ =\dfrac{2x^2 +2x-2x+2-4}{(x-1)(x+1)} \\ = \dfrac{2x^2 -2}{(x-1)(x+1)} \\ = \dfrac{2(x^2-1)}{x^2-1} =\tt 2 [/tex]
Punctul b)
[tex] \dfrac{x+2}{x-2} -\dfrac{x-2}{x+2} -\dfrac{16}{x^2-4} \\ = \dfrac{x+2}{x-2} -\dfrac{x-2}{x+2} -\dfrac{16}{(x-2)(x+2)} \\ = \dfrac{(x+2)^2 -(x-2)^2 -16}{(x-2)(x+2)} \\ =\dfrac{x^2 +4x+4 -(x^2 -4x+4) -16}{(x-2)( x+2)} \\ =\dfrac{ 8x-16 }{(x-2)(x+2)} \\ =\dfrac{8(x-2)}{(x-2)(x+2)} =\tt \dfrac{8}{x+2} [/tex]
Punctul c)
[tex] \dfrac{x-1}{x-2} -\dfrac{x+3}{x+2} +\dfrac{2}{x^2 -4} \\ = \dfrac{x-1}{x-2} -\dfrac{x+3}{x+2} +\dfrac{2}{(x-2)(x+2)} \\ = \dfrac{(x-1)(x+2)-(x+3)(x-2)+2}{(x-2)(x+2)} \\ =\dfrac{x^2 +x-2-x^2-x+6+2}{(x-2)(x+2)} \\ = \tt \dfrac{6}{(x-2)(x+2)}[/tex]
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