Question

\frac{2}{x^{2}-4}+\frac{1}{2x-x^{2}}: \frac{1}{x^{2}+4x+4}


Упростить

Answer 1

$(\frac{2}{x^{2}-4}+\frac{1}{2x-x^{2}}): \frac{1}{x^{2}+4x+4} = \\ = (\frac{2}{(x-2)(x + 2)}+\frac{1}{ - x( - 2 + x)}): \frac{1}{(x + 2)^{2}} = \\ = (\frac{2}{(x-2)(x + 2)} - \frac{1}{ x(x - 2)} )\times \frac{(x + 2)^{2}}{1} = \\ = (\frac{2х}{х(x-2)(x + 2)} - \frac{х+2}{ x(x - 2)(х+2)} )\times \frac{(x + 2)^{2}}{1} =\\ =\frac{2х-х-2}{х(x-2)(x + 2)} \times \frac{(x + 2)^{2}}{1}=\\=\frac{х-2}{х(x-2)(x + 2)} \times \frac{(x + 2)^{2}}{1}=\\=\frac{1}{х} \times \frac{x + 2}{1}=\\=\frac{x + 2}{х}$