Question

Aratati ca E(x)=$( \frac{x}{x+2} + \frac{1}{ x^{2} -4} )( \frac{x+1}{x-1} + \frac{2x+5}{1- x^{2} } ):( \frac{1}{x+1} - \frac{1}{x+1} - \frac{1}{2x} ) =2x$ ,une x apartine R\{-2,-1,0,1,2}

Answer 1

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Answer 2

$E(x)= [\frac{x}{x+2}+ \frac{1}{(x+2)(x-2)}] *[ \frac{x+1}{x-1}+ \frac{2x+5}{(1-x)(1+x)}] :( \frac{1}{x+1}- \frac{1}{2x})= \\ = \frac{x(x-2)+1}{(x+2)(x-2)}* [ \frac{x+1}{x-1}- \frac{2x+5}{(x-1)(1+x)}]:[ \frac{2x-(x+1)}{2x(x+1)}]= \\ = \frac{ x^{2} -2x+1}{(x+2)(x-2)}* \frac{(x+1)^{2}-(2x+5)}{(x-1)(x+1)}: \frac{x-1}{2x(x+1)}= \\ = \frac{(x-1) ^{2} }{(x+2)(x-2)}* \frac{ x^{2} +2x+1-2x-5}{(x-1)(x+1)}* \frac{2x(x+1)}{x-1} = \\ = \frac{(x-1) ^{2}*( x^{2} -4)*2x(x+1) }{(x-1)^{2}(x+2)(x-2)(x+1)}=$
$= \frac{( x^{2} -4)*2x}{(x+2)(x-2)}= \\ = \frac{(x+2)(x-2)*2x}{(x+2)(x-2)} = \\ =\boxed{2x}.$