Tính:

\(\frac{{1 - 4{x^2}}}{{{x^2} + 4x}}:\frac{{2 - 4x}}{{3x}}\).

P = \(\left( {\frac{1}{{x - 1}} - \frac{x}{{1 - {x^3}}}.\frac{{{x^2} + x + 1}}{{x + 1}}} \right):\frac{{2x + 1}}{{{x^2} + 2x + 1}}\)

\( = \left[ {\frac{1}{{x - 1}} - \frac{x}{{\left( {1 - x} \right)\left( {{x^2} + x + 1} \right)}}.\frac{{{x^2} + x + 1}}{{x + 1}}} \right]:\frac{{2x + 1}}{{{{\left( {x + 1} \right)}^2}}}\)

\( = \left[ {\frac{1}{{x - 1}} + \frac{{x\left( {{x^2} + x + 1} \right)}}{{\left( {x - 1} \right)\left( {{x^2} + x + 1} \right)\left( {x + 1} \right)}}} \right]:\frac{{2x + 1}}{{{{\left( {x + 1} \right)}^2}}}\)

\( = \left[ {\frac{1}{{x - 1}} + \frac{x}{{\left( {x - 1} \right)\left( {x + 1} \right)}}} \right]:\frac{{2x + 1}}{{{{\left( {x + 1} \right)}^2}}}\)

\( = \left[ {\frac{{x + 1}}{{\left( {x - 1} \right)\left( {x + 1} \right)}} + \frac{x}{{\left( {x - 1} \right)\left( {x + 1} \right)}}} \right]:\frac{{2x + 1}}{{{{\left( {x + 1} \right)}^2}}}\)

\( = \left[ {\frac{{x + 1 + x}}{{\left( {x - 1} \right)\left( {x + 1} \right)}}} \right]:\frac{{2x + 1}}{{{{\left( {x + 1} \right)}^2}}}\)

\( = \frac{{2x + 1}}{{\left( {x - 1} \right)\left( {x + 1} \right)}}.\frac{{{{\left( {x + 1} \right)}^2}}}{{2x + 1}}\)\( = \frac{{x + 1}}{{x - 1}}\).

Điều kiện xác định của P là: \(\left\{ \begin{array}{l}x - 1 \ne 0\\1 - {x^3} \ne 0\\x + 1 \ne 0\\{x^2} + 2x + 1 = {\left( {x + 1} \right)^2} \ne 0\end{array} \right.hay\,\,\left\{ \begin{array}{l}x \ne 1\\x \ne - 1\end{array} \right.\).