Đáp án:
$\begin{array}{l}
a)\,\,{\frac{{ - 2}}{{x - 2}}}\\
b)\,\,\,\frac{{x + 1}}{{{x^2} + x + 1}}
\end{array}$
Giải thích các bước giải:
\(\begin{array}{*{20}{l}}
{a){\kern 1pt} {\kern 1pt} {\kern 1pt} \frac{1}{{x + 2}} + \frac{2}{{2 - x}} + \frac{x}{{{x^2} - 4}}}\\
{\,\,\, = \frac{1}{{x + 2}} - \frac{2}{{x - 2}} + \frac{x}{{\left( {x - 2} \right)\left( {x + 2} \right)}}}\\
{{\rm{}} = \frac{{x - 2 - 2\left( {x + 2} \right) + x}}{{\left( {x - 2} \right)\left( {x + 2} \right)}} = \frac{{x - 2x - 2x - 4 + x}}{{\left( {x - 2} \right)\left( {x + 2} \right)}}}\\
{{\rm{}} = \frac{{ - 2\left( {x + 2} \right)}}{{\left( {x - 2} \right)\left( {x + 2} \right)}} = \frac{{ - 2}}{{x - 2}}.}\\
{b){\kern 1pt} {\kern 1pt} \frac{{3{x^2} + 5x + 1}}{{{x^3} - 1}} - \frac{{1 - x}}{{{x^2} + x + 1}} - \frac{3}{{x - 1}}}\\
{n{\rm{}} = \frac{{3{x^2} + 5x + 1}}{{\left( {x - 1} \right)\left( {{x^2} + x + 1} \right)}} + \frac{{x - 1}}{{{x^2} + x + 1}} - \frac{3}{{x - 1}}}\\
{n{\rm{}} = \frac{{3{x^2} + 5x + 1 + {{\left( {x - 1} \right)}^2} - 3\left( {{x^2} + x + 1} \right)}}{{\left( {x - 1} \right)\left( {{x^2} + x + 1} \right)}}}\\
{n{\rm{}} = \frac{{3{x^2} + 5x + 1 + {x^2} - 2x + 1 - 3{x^2} - 3x - 3}}{{\left( {x - 1} \right)\left( {{x^2} + x + 1} \right)}}}\\
{n{\rm{}} = \frac{{{x^2} - 1}}{{\left( {x - 1} \right)\left( {{x^2} + x + 1} \right)}} = \frac{{\left( {x - 1} \right)\left( {x + 1} \right)}}{{\left( {x - 1} \right)\left( {{x^2} + x + 1} \right)}}}\\
{n{\rm{}} = \frac{{x + 1}}{{{x^2} + x + 1}}.}
\end{array}\)
