Giải thích các bước giải:
2.
$\begin{gathered} a)\,\frac{{{x^2}}}{{x - 1}} + \frac{1}{{x + 1}} - \frac{{2x}}{{{x^2} - 1}} \hfill \\ = \frac{{{x^2}(x + 1) + x - 1 - 2x}}{{(x - 1)(x + 1)}} \hfill \\ = \frac{{{x^3} + {x^2} - x - 1}}{{(x - 1)(x + 1)}} \hfill \\ = \frac{{{x^2}(x + 1) - (x + 1)}}{{(x - 1)(x + 1)}} \hfill \\ = \frac{{(x + 1)({x^2} - 1)}}{{(x - 1)(x + 1)}} \hfill \\ = \frac{{(x - 1){{(x + 1)}^2}}}{{(x - 1)(x + 1)}} \hfill \\ = x + 1 \hfill \\ \end{gathered} $
$\begin{gathered} b)\,(x - \frac{{{x^2} + {y^2}}}{{x + y}}).(\frac{1}{y} + \frac{2}{{x - y}}) \hfill \\ = (\frac{{x(x + y) - {x^2} - {y^2}}}{{x + y}}).(\frac{{x - y + 2y}}{{y(x - y)}}) \hfill \\ = \frac{{{x^2} + xy - {x^2} - {y^2}}}{{x + y}}.\frac{{x + y}}{{y(x - y)}} \hfill \\ = \frac{{y(x - y)(x + y)}}{{(x + y).y(x - y)}} \hfill \\ = 1 \hfill \\ \end{gathered} $
$\begin{gathered} c)\,(\frac{x}{{{x^2} - 36}} - \frac{{x - 6}}{{{x^2} + 6x}}):\frac{{2x - 6}}{{{x^2} + 6x}} + \frac{x}{{6 - x}} \hfill \\ = \frac{{{x^2} - {{(x - 6)}^2}}}{{x(x - 6)(x + 6)}}.\frac{{x(x + 6)}}{{2(x - 3)}} + \frac{x}{{6 - x}} \hfill \\ = \frac{{\left[ {x - (x - 6)} \right]\left[ {x + x - 6} \right]}}{{x - 6}}.\frac{1}{{2(x - 3)}} - \frac{x}{{x - 6}} \hfill \\ = \frac{{6.2(x - 3)}}{{x - 6}}.\frac{1}{{2(x - 3)}} - \frac{x}{{x - 6}} \hfill \\ = \frac{6}{{x - 6}} - \frac{x}{{x - 6}} \hfill \\ = \frac{{6 - x}}{{x - 6}} = - 1 \hfill \\ \end{gathered} $
