Đáp án:
$\begin{array}{l}
b)B = {x^2}\left( {{y^2} - x{y^2} + 1} \right) - {y^2}\left( {{x^3} + {x^2} + 1} \right)\\
= {x^2}{y^2} - {x^3}{y^2} + {x^2} - {x^3}{y^2} - {x^2}{y^2} - {y^2}\\
= - 2{x^3}{y^2} + {x^2} - {y^2}\\
= - 2.{\left( {\frac{1}{2}} \right)^3}.{\left( {\frac{1}{2}} \right)^2} + {\left( {\frac{1}{2}} \right)^2} - {\left( {\frac{{ - 1}}{2}} \right)^2}\\
= - 2.\frac{1}{8}.\frac{1}{4}\\
= \frac{{ - 1}}{{16}}\\
c)C = {x^6} - 20{x^5} - 20{x^4}\\
- 20{x^3} - 20{x^2} - 20x + 3\\
= {x^6} - 21{x^5} + {x^5} - 21{x^4} + {x^4} - 21{x^3}\\
+ {x^3} - 21{x^2} + {x^2} - 21x + x + 3\\
= {x^5}\left( {x - 21} \right) + {x^4}\left( {x - 21} \right) + {x^3}\left( {x - 21} \right)\\
+ {x^2}\left( {x - 21} \right) + x\left( {x - 21} \right) + x + 3\\
= \left( {x - 21} \right)\left( {{x^5} + {x^4} + {x^3} + {x^2} + x} \right) + x + 3\\
= \left( {21 - 21} \right).\left( {{x^5} + {x^4} + {x^3} + {x^2} + x} \right) + 21 + 3\\
= 0 + 21 + 3\\
= 24
\end{array}$
