Đáp án:

 b) -8

Giải thích các bước giải:

\(\begin{array}{l}
a)\dfrac{{11x + 13}}{{3\left( {x - 1} \right)}} + \dfrac{{15x + 17}}{{4\left( {1 - x} \right)}}\\
 = \dfrac{{4\left( {11x + 13} \right) - 3\left( {15x + 17} \right)}}{{12\left( {x - 1} \right)}}\\
 = \dfrac{{ - x + 1}}{{12\left( {x - 1} \right)}} =  - \dfrac{1}{{12}}\\
b)\dfrac{{2x + 1}}{{2{x^2} - x}} + \dfrac{{32{x^2}}}{{1 - 4{x^2}}} + \dfrac{{1 - 2x}}{{2{x^2} + x}}\\
 = \dfrac{{2x + 1}}{{x\left( {2x - 1} \right)}} + \dfrac{{32{x^2}}}{{\left( {1 - 2x} \right)\left( {1 + 2x} \right)}} + \dfrac{{1 - 2x}}{{x\left( {2x + 1} \right)}}\\
 = \dfrac{{ - {{\left( {2x + 1} \right)}^2} + 32{x^3} + {{\left( {1 - 2x} \right)}^2}}}{{x\left( {1 - 2x} \right)\left( {1 + 2x} \right)}}\\
 = \dfrac{{ - 4{x^2} - 4x - 1 + 32{x^3} + 1 - 4x + 4{x^2}}}{{x\left( {1 - 2x} \right)\left( {1 + 2x} \right)}}\\
 = \dfrac{{32{x^3} - 8x}}{{x\left( {1 - 2x} \right)\left( {1 + 2x} \right)}}\\
 = \dfrac{{8\left( {4{x^2} - 1} \right)}}{{1 - 4{x^2}}} =  - 8\\
c)\dfrac{1}{{{x^2} + x + 1}} + \dfrac{1}{{{x^2} - x}} + \dfrac{1}{{1 - {x^3}}}\\
 = \dfrac{1}{{{x^2} + x + 1}} + \dfrac{1}{{x\left( {x - 1} \right)}} - \dfrac{1}{{\left( {x - 1} \right)\left( {{x^2} + x + 1} \right)}}\\
 = \dfrac{{x\left( {x - 1} \right) + {x^2} + x + 1 - x}}{{x\left( {x - 1} \right)\left( {{x^2} + x + 1} \right)}}\\
 = \dfrac{{2{x^2} - x + 1}}{{x\left( {x - 1} \right)\left( {{x^2} + x + 1} \right)}}\\
d)\dfrac{{x + 9}}{{{x^2} - 9}} - \dfrac{3}{{{x^2} + 3x}}\\
 = \dfrac{{x\left( {x + 9} \right) - 3\left( {x - 3} \right)}}{{x\left( {x + 3} \right)\left( {x - 3} \right)}}\\
 = \dfrac{{{x^2} + 9x - 3x + 9}}{{x\left( {x + 3} \right)\left( {x - 3} \right)}}\\
 = \dfrac{{{x^2} + 6x + 9}}{{x\left( {x + 3} \right)\left( {x - 3} \right)}} = \dfrac{{{{\left( {x + 3} \right)}^2}}}{{x\left( {x + 3} \right)\left( {x - 3} \right)}}\\
 = \dfrac{{x + 3}}{{x\left( {x - 3} \right)}}
\end{array}\)