Đáp án:
$\begin{array}{l}
a)\left( {4{x^2} + 2x + 1} \right)\left( {2x - 1} \right)\\
= {\left( {2x} \right)^3} - {1^3}\\
= 8{x^3} - 1\\
b)\frac{x}{{2x - 2}} + \frac{{ - {x^2} - 1}}{{2 - 2{x^2}}} - 1\\
= \frac{x}{{2\left( {x - 1} \right)}} + \frac{{{x^2} + 1}}{{2\left( {x - 1} \right)\left( {x + 1} \right)}} - 1\\
= \frac{{x\left( {x + 1} \right) + {x^2} + 1 - \left( {2 - 2{x^2}} \right)}}{{2\left( {x - 1} \right)\left( {x + 1} \right)}}\\
= \frac{{{x^2} + x + {x^2} + 1 - 2 + 2{x^2}}}{{2\left( {x - 1} \right)\left( {x + 1} \right)}}\\
= \frac{{4{x^2} + x - 1}}{{2 - 2{x^2}}}\\
c)\left( {3{x^4} + {x^3} - 7{x^2} - 2x + 2} \right):\left( {3{x^2} + x - 1} \right)\\
= {x^2} - 2\\
d)\left( {x - 2y} \right)\left( {4{x^2} + 2xy + {y^2}} \right)\\
= 4{x^3} + 2{x^2}y + x{y^2} - 8{x^2}y - 4x{y^2} - 2{y^3}\\
= 4{x^3} - 6{x^2}y - 3x{y^2} - 2{y^3}\\
e)\frac{{ - {a^2} + {b^2}}}{{3ab}} - \frac{{3a - 2b}}{{6b}} - \frac{b}{{12a}}\\
= \frac{{ - 4{a^2} + 4{b^2} - 2a.\left( { - 3a - 2b} \right) - b.b}}{{12ab}}\\
= \frac{{ - 4{a^2} + 4{b^2} + 6{a^2} + 4ab - {b^2}}}{{12ab}}\\
= \frac{{2{a^2} + 4ab + 3{b^2}}}{{12ab}}\\
f)\frac{{x + 1}}{{2x - 2}} - \frac{{{x^2} + 3}}{{2{x^2} - 2}}\\
= \frac{{\left( {x + 1} \right).\left( {x + 1} \right) - {x^2} - 3}}{{2\left( {x - 1} \right)\left( {x + 1} \right)}}\\
= \frac{{{x^2} + 2x + 1 - {x^2} - 3}}{{2\left( {x - 1} \right)\left( {x + 1} \right)}}\\
= \frac{{2x - 2}}{{2\left( {x - 1} \right)\left( {x + 1} \right)}}\\
= \frac{1}{{x + 1}}
\end{array}$
