Đáp án:
$\begin{array}{l}
Dkxd:x \ne 0;x \ne - 1\\
M = \left( {\dfrac{{x + 2}}{{3x}} + \dfrac{2}{{x + 1}} - 3} \right):\dfrac{{2 - 4x}}{{x + 1}}\\
- \dfrac{{3x - {x^2} + 1}}{{3x}}\\
= \dfrac{{\left( {x + 2} \right).\left( {x + 1} \right) + 2.3x - 3.3x\left( {x + 1} \right)}}{{3x\left( {x + 1} \right)}}\\
.\dfrac{{x + 1}}{{2 - 4x}} - \dfrac{{3x - {x^2} + 1}}{{3x}}\\
= \dfrac{{{x^2} + 3x + 2 + 6x - 9{x^2} - 9x}}{{3x}}.\dfrac{1}{{2 - 4x}}\\
- \dfrac{{3x - {x^2} + 1}}{{3x}}\\
= \dfrac{{ - 8{x^2} + 2}}{{3x}}.\dfrac{1}{{2.\left( {1 - 2x} \right)}} - \dfrac{{3x - {x^2} + 1}}{{3x}}\\
= \dfrac{{2.\left( {1 - 4{x^2}} \right)}}{{6x\left( {1 - 2x} \right)}} - \dfrac{{3x - {x^2} + 1}}{{3x}}\\
= \dfrac{{\left( {1 - 2x} \right)\left( {1 + 2x} \right)}}{{3x\left( {1 - 2x} \right)}} - \dfrac{{3x - {x^2} + 1}}{{3x}}\\
= \dfrac{{1 + 2x}}{{3x}} - \dfrac{{3x - {x^2} + 1}}{{3x}}\\
= \dfrac{{1 + 2x - 3x + {x^2} - 1}}{{3x}}\\
= \dfrac{{{x^2} - x}}{{3x}}\\
= \dfrac{{x - 1}}{3}
\end{array}$
