Đáp án:
$\begin{array}{l}
b)\frac{{4x + 12}}{{3{x^2}.x}}:\frac{{{x^3} + 3x}}{{1 - 3x}}\\
= \frac{{4\left( {x + 3} \right)}}{{3{x^3}}}:\frac{{x\left( {{x^2} + 3} \right)}}{{1 - 3x}}\\
= \frac{{4\left( {x + 3} \right)}}{{3{x^3}}}.\frac{{1 - 3x}}{{x\left( {{x^2} + 3} \right)}}
\end{array}$
=> đề bài có vấn đề
$\begin{array}{l}
c)\frac{{x + 1}}{{{x^2}\left( {x - 1} \right)}} - \frac{{x - 2}}{{8x\left( {{x^2} - 1} \right)}}\\
= \frac{{x + 1}}{{{x^2}\left( {x - 1} \right)}} - \frac{{x - 2}}{{8x\left( {x - 1} \right)\left( {x + 1} \right)}}\\
= \frac{{\left( {x + 1} \right).8\left( {x + 1} \right) - x\left( {x - 2} \right)}}{{8{x^2}\left( {x - 1} \right)\left( {x + 1} \right)}}\\
= \frac{{8\left( {{x^2} + 2x + 1} \right) - {x^2} + 2x}}{{8{x^2}\left( {x + 1} \right)\left( {x - 1} \right)}}\\
= \frac{{7{x^2} + 18x + 8}}{{8{x^2}\left( {x - 1} \right)\left( {x + 1} \right)}}\\
= \frac{{\left( {x + 2} \right)\left( {7x + 4} \right)}}{{8{x^2}\left( {x - 1} \right)\left( {x + 1} \right)}}\\
d)\left( {\frac{1}{{1 + x}} + \frac{{2x}}{{1 - {x^2}}}} \right):\left( {\frac{1}{x} - 1} \right)\\
= \frac{{1 - x + 2x}}{{\left( {1 - x} \right)\left( {1 + x} \right)}}:\frac{{1 - x}}{x}\\
= \frac{{1 + x}}{{\left( {1 - x} \right)\left( {1 + x} \right)}}.\frac{x}{{1 - x}}\\
= \frac{x}{{{{\left( {1 - x} \right)}^2}}}\\
e)\left( {\frac{{x + 1}}{{x - 1}} - \frac{1}{{{x^2} - 1}}} \right).\frac{{x + 1}}{{x + 2}}\\
= \frac{{\left( {x + 1} \right)\left( {x + 1} \right) - 1}}{{\left( {x - 1} \right)\left( {x + 1} \right)}}.\frac{{x + 1}}{{x + 2}}\\
= \frac{{{x^2} + 2x + 2}}{{\left( {x - 1} \right)\left( {x + 2} \right)}}
\end{array}$
$\begin{array}{l}
\left( {\frac{{3x}}{{1 - 3x}} + \frac{{2x}}{{3x + 1}}} \right):\frac{{6{x^2} + 10x}}{{1 - 6x + 9{x^2}}}\\
= \frac{{3x\left( {3x + 1} \right) + 2x\left( {1 - 3x} \right)}}{{1 - 9{x^2}}}.\frac{{{{\left( {3x - 1} \right)}^2}}}{{2x\left( {3x + 5} \right)}}\\
= \frac{{9{x^2} + 3x + 2x - 6{x^2}}}{{\left( {3x + 1} \right)}}.\frac{{1 - 3x}}{{2x\left( {3x + 5} \right)}}\\
= \frac{{3{x^2} + 5x}}{{\left( {3x + 1} \right)}}.\frac{{1 - 3x}}{{2x\left( {3x + 5} \right)}}\\
= \frac{{1 - 3x}}{{2\left( {3x + 1} \right)}}\\
= \frac{{1 - 3x}}{{6x + 2}}
\end{array}$
