Đáp án:
$\begin{array}{l}
1)A = {x^{15}} + 3{x^{14}} + 5\\
= {x^{14}}\left( {x + 3} \right) + 5\\
= {x^{14}}.0 + 5\\
= 5\\
2)B = {\left( {{x^{2007}} + 3{x^{2006}} + 1} \right)^{2007}}\\
= {\left[ {{x^{2016}}.\left( {x + 3} \right) + 1} \right]^{2007}}\\
= {\left( {{x^{2016}}.\left( { - 3 + 3} \right) + 1} \right)^{2007}}\\
= {1^{2007}}\\
= 1\\
3)C = 21{x^4} + 12{x^3} - 3{x^2} + 24x + 15\\
= 3x\left( {7{x^3} + 4{x^2} - x + 8} \right) + 15\\
= 3x.0 + 15\\
= 15\\
4)D = - 16{x^5} - 28{x^4} + 16{x^3} - 20{x^2} + 32x + 2007\\
= 4x\left( { - 4{x^4} - 7{x^3} + 4{x^2} - 5x + 8} \right) + 2007\\
= 4x.0 + 2007\\
= 2007
\end{array}$
