$$ \dfrac{3x+6}{x^2+4x+4}÷\dfrac{x-2}{x+2} -\dfrac{1}{x-2} $$ , in which
$$ x=-6 $$

Question

$$ \dfrac{3x+6}{x^2+4x+4}÷\dfrac{x-2}{x+2}  -\dfrac{1}{x-2}  $$ , in which
$$ x=-6 $$

Uniteasy Admin · Accepted Answer

First simplify the expression
$$ \dfrac{3x+6}{x^2+4x+4}÷\dfrac{x-2}{x+2} -\dfrac{1}{x-2} $$
$$ =\dfrac{3(x+2)}{(x+2)^2}\cdotp \dfrac{x+2}{x-2} -\dfrac{1}{x-2}  $$
$$ =\dfrac{3}{x-2}  -\dfrac{1}{x-2} =\dfrac{2}{x-2}=-\dfrac{1}{4} $$

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