Giải thích các bước giải:

B7:

a) ĐKXĐ: $x\ne \{\pm 2\}$

$\begin{array}{l}
\dfrac{4}{{x + 2}} + \dfrac{2}{{x - 2}} + \dfrac{{5x - 6}}{{4 - {x^2}}}\\
 = \dfrac{4}{{x + 2}} + \dfrac{2}{{x - 2}} + \dfrac{{5x - 6}}{{\left( {2 - x} \right)\left( {2 + x} \right)}}\\
 = \dfrac{{4\left( {x - 2} \right) + 2\left( {x + 2} \right) + 6 - 5x}}{{\left( {x - 2} \right)\left( {x + 2} \right)}}\\
 = \dfrac{{x + 2}}{{\left( {x - 2} \right)\left( {x + 2} \right)}}\\
 = \dfrac{1}{{x - 2}}
\end{array}$

b) ĐKXĐ: $x\ne \{0;\dfrac{1}{2};2\}$

$\begin{array}{l}
\dfrac{{1 - 3x}}{{2 - x}} + \dfrac{{3x - 2}}{{2x - 1}} + \dfrac{{3x + 2}}{{2x - 4{x^2}}}\\
 = \dfrac{{1 - 3x}}{{2 - x}} + \dfrac{{3x - 2}}{{2x - 1}} + \dfrac{{3x + 2}}{{2x\left( {1 - 2x} \right)}}\\
 = \dfrac{{\left( {3x - 1} \right)2x\left( {2x - 1} \right) + \left( {3x - 2} \right)2x\left( {x - 2} \right) + \left( {3x + 2} \right)\left( {2 - x} \right)}}{{\left( {x - 2} \right)2x\left( {2x - 1} \right)}}\\
 = \dfrac{{18{x^3} - 29{x^2} + 14x + 4}}{{\left( {x - 2} \right)2x\left( {2x - 1} \right)}}
\end{array}$

B8:

a) ĐKXĐ: $x\ne \{\pm 3\}$

$\begin{array}{l}
\dfrac{1}{{{x^2} + 6x + 9}} + \dfrac{1}{{6x - {x^2} - 9}} + \dfrac{x}{{{x^2} - 9}}\\
 = \dfrac{1}{{{{\left( {x + 3} \right)}^2}}} - \dfrac{1}{{{{\left( {x - 3} \right)}^2}}} + \dfrac{x}{{\left( {x - 3} \right)\left( {x + 3} \right)}}\\
 = \dfrac{{{{\left( {x - 3} \right)}^2} - {{\left( {x + 3} \right)}^2} + x\left( {x - 3} \right)\left( {x + 3} \right)}}{{{{\left( {x - 3} \right)}^2}{{\left( {x + 3} \right)}^2}}}\\
 = \dfrac{{x\left( {{x^2} - 21} \right)}}{{{{\left( {x - 3} \right)}^2}{{\left( {x + 3} \right)}^2}}}
\end{array}$

b) ĐKXĐ: $x\ne 1$

$\begin{array}{l}
\dfrac{{{x^2} + 2}}{{{x^3} + 1}} + \dfrac{2}{{{x^2} + x + 1}} + \dfrac{1}{{1 - x}}\\
 = \dfrac{{{x^2} + 2}}{{\left( {{x^2} + x + 1} \right)\left( {x - 1} \right)}} + \dfrac{2}{{{x^2} + x + 1}} + \dfrac{1}{{1 - x}}\\
 = \dfrac{{{x^2} + 2 + 2\left( {x - 1} \right) - \left( {{x^2} + x + 1} \right)}}{{\left( {{x^2} + x + 1} \right)\left( {x - 1} \right)}}\\
 = \dfrac{{x - 1}}{{\left( {{x^2} + x + 1} \right)\left( {x - 1} \right)}}\\
 = \dfrac{1}{{{x^2} + x + 1}}
\end{array}$