Đáp án:
$\begin{array}{l}
a)\dfrac{{x + 1}}{{x + 2}}:\left( {\dfrac{{x + 2}}{{x + 3}}:\dfrac{{x + 3}}{{x + 1}}} \right)\\
= \dfrac{{x + 1}}{{x + 2}}:\left( {\dfrac{{x + 2}}{{x + 3}}.\dfrac{{x + 1}}{{x + 3}}} \right)\\
= \dfrac{{x + 1}}{{x + 2}}:\dfrac{{\left( {x + 2} \right)\left( {x + 1} \right)}}{{{{\left( {x + 3} \right)}^2}}}\\
= \dfrac{{x + 1}}{{x + 2}}.\dfrac{{{{\left( {x + 3} \right)}^2}}}{{\left( {x + 2} \right)\left( {x + 1} \right)}}\\
= \dfrac{{{{\left( {x + 3} \right)}^2}}}{{{{\left( {x + 2} \right)}^2}}}\\
b)\left( {\dfrac{{3x}}{{1 - 3x}} + \dfrac{{2x}}{{3x + 1}}} \right):\dfrac{{6{x^2} + 10x}}{{1 - 6x + 9{x^2}}}\\
= \dfrac{{3x\left( {3x + 1} \right) + 2x\left( {1 - 3x} \right)}}{{\left( {1 - 3x} \right)\left( {3x + 1} \right)}}.\dfrac{{{{\left( {1 - 3x} \right)}^2}}}{{2x\left( {3x + 5} \right)}}\\
= \dfrac{{9{x^2} + 3x + 2x - 6{x^2}}}{{3x + 1}}.\dfrac{{1 - 3x}}{{2x\left( {3x + 5} \right)}}\\
= \dfrac{{x\left( {3x + 5} \right)}}{{3x + 1}}.\dfrac{{1 - 3x}}{{2x\left( {3x + 5} \right)}}\\
= \dfrac{{1 - 3x}}{{2\left( {3x + 1} \right)}}\\
c)\left( {\dfrac{1}{{{x^2} + x}} - \dfrac{{2 - x}}{{x + 1}}} \right):\left( {\dfrac{1}{x} + x - 2} \right)\\
= \dfrac{{1 - x.\left( {2 - x} \right)}}{{x\left( {x + 1} \right)}}:\dfrac{{1 + x\left( {x - 2} \right)}}{x}\\
= \dfrac{{1 - 2x + {x^2}}}{{x\left( {x + 1} \right)}}.\dfrac{x}{{1 + {x^2} - 2x}}\\
= \dfrac{1}{{x + 1}}\\
d)\left( {\dfrac{9}{{{x^3} - 9x}} + \dfrac{1}{{x + 3}}} \right):\left( {\dfrac{{x - 3}}{{{x^2} + 3x}} - \dfrac{x}{{3x + 9}}} \right)\\
= \dfrac{{9 + x\left( {x - 3} \right)}}{{x\left( {x + 3} \right)\left( {x - 3} \right)}}:\dfrac{{3.\left( {x - 3} \right) - x.x}}{{3x\left( {x + 3} \right)}}\\
= \dfrac{{{x^2} - 3x + 9}}{{x\left( {x + 3} \right)\left( {x - 3} \right)}}.\dfrac{{3x\left( {x + 3} \right)}}{{3x - 9 - {x^2}}}\\
= \dfrac{3}{{x - 3}}
\end{array}$
