Answer
$\dfrac{(x+2)(x-1)}{x(x+1)}$
Work Step by Step
The given expression, $
\dfrac{2+\dfrac{2}{1+x}}{2-\dfrac{2}{1-x}}
,$ simplifies to
\begin{array}{l}\require{cancel}
\dfrac{\dfrac{2(1+x)+2}{1+x}}{\dfrac{2(1-x)-2}{1-x}}
\\\\=
\dfrac{\dfrac{2+2x+2}{1+x}}{\dfrac{2-2x-2}{1-x}}
\\\\=
\dfrac{\dfrac{2x+4}{1+x}}{\dfrac{-2x}{1-x}}
\\\\=
\dfrac{2x+4}{1+x}\div\dfrac{-2x}{1-x}
\\\\=
\dfrac{2x+4}{1+x}\cdot\dfrac{1-x}{-2x}
\\\\=
\dfrac{2(x+2)}{1+x}\cdot\dfrac{1-x}{-2x}
\\\\=
\dfrac{\cancel{2}(x+2)}{1+x}\cdot\dfrac{1-x}{-\cancel{2}x}
\\\\=
\dfrac{x+2}{1+x}\cdot\dfrac{1-x}{-x}
\\\\=
\dfrac{(x+2)(1-x)}{-x(1+x)}
\\\\=
-\dfrac{(x+2)(1-x)}{x(x+1)}
\\\\=
\dfrac{(x+2)(x-1)}{x(x+1)}
.\end{array}
