Đáp án:

\(\begin{array}{l}
f)\dfrac{{90x + 2}}{{1 - {x^2}}}\\
g)\dfrac{{12{x^2} + 6x}}{{\left( {2x + 2} \right)\left( {2x - 1} \right)}}
\end{array}\)

Giải thích các bước giải:

\(\begin{array}{l}
f)\dfrac{6}{{1 + x}} - \dfrac{4}{{1 - x}} - \dfrac{{100x}}{{{x^2} - 1}}\\
 = \dfrac{{6\left( {1 - x} \right) - 4\left( {x + 1} \right) + 100x}}{{\left( {1 - x} \right)\left( {x + 1} \right)}}\\
 = \dfrac{{6 - 6x - 4x - 4 + 100x}}{{\left( {1 - x} \right)\left( {x + 1} \right)}}\\
 = \dfrac{{90x + 2}}{{1 - {x^2}}}\\
g)\dfrac{{2x + 1}}{{2x + 2}} - \dfrac{{2x - 2}}{{2x - 1}} + \dfrac{3}{2}\\
 = \dfrac{{\left( {2x + 1} \right)\left( {2x - 1} \right) - \left( {2x - 2} \right)\left( {2x + 2} \right)}}{{\left( {2x + 2} \right)\left( {2x - 1} \right)}} + \dfrac{3}{2}\\
 = \dfrac{{4{x^2} - 1 - 4{x^2} + 4}}{{\left( {2x + 2} \right)\left( {2x - 1} \right)}} + \dfrac{3}{2}\\
 = \dfrac{3}{{\left( {2x + 2} \right)\left( {2x - 1} \right)}} + \dfrac{3}{2}\\
 = \dfrac{{6 + 3\left( {4{x^2} + 2x - 2} \right)}}{{2\left( {2x + 2} \right)\left( {2x - 1} \right)}}\\
 = \dfrac{{12{x^2} + 6x}}{{\left( {2x + 2} \right)\left( {2x - 1} \right)}}
\end{array}\)