\begin{array}{l} a)A = \left( {\dfrac{1}{{x - 1}} - \dfrac{x}{{1 - {x^2}}}.\dfrac{{{x^2} + x + 1}}{{x + 1}}} \right):\dfrac{{2x + 1}}{{{x^2} + 2x + 1}}\\ = \left[ {\dfrac{{{x^2} + 2x + 1 + x\left( {{x^2} + x + 1} \right)}}{{\left( {x - 1} \right){{\left( {x + 1} \right)}^2}}}} \right].\dfrac{{{{\left( {x + 1} \right)}^2}}}{{2x + 1}}\\ = \dfrac{{{x^2} + 2x + 1 + {x^3} + {x^2} + x}}{{x{{\left( {x + 1} \right)}^2}}}.\dfrac{{{{\left( {x + 1} \right)}^2}}}{{2x + 1}}\\ = \dfrac{{{x^3} + 2{x^2} + 3x + 1}}{{x\left( {2x + 1} \right)}}\\ b)Thay:x = \dfrac{1}{2}\\ \to A = \dfrac{{{{\left( {\dfrac{1}{2}} \right)}^3} + 2.{{\left( {\dfrac{1}{2}} \right)}^2} + 3.\left( {\dfrac{1}{2}} \right) + 1}}{{\dfrac{1}{2}.\left( {2.\dfrac{1}{2} + 1} \right)}} = \dfrac{{25}}{8} \end{array}
