Đáp án:
2) 8xy
Giải thích các bước giải:
\(\begin{array}{l}
1){x^3} + 12x + 3 \vdots x + 2\\
\Leftrightarrow {x^3} + 2{x^2} - 2{x^2} - 4x + 16x + 32 - 29 \vdots x + 2\\
\to {x^2}\left( {x + 2} \right) - 2x\left( {x + 2} \right) + 16\left( {x + 2} \right) - 29 \vdots x + 2\\
\to 29 \vdots x + 2\\
\to x + 2 \in U\left( {29} \right)\\
\to \left[ \begin{array}{l}
x + 2 = 29\\
x + 2 = - 29\\
x + 2 = 1\\
x + 2 = - 1
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = 27\\
x = - 31\\
x = - 1\\
x = - 3
\end{array} \right.\\
2){\left( {2x + y} \right)^2} - {\left( {2x - y} \right)^2}\\
= \left( {2x + y - 2x + y} \right)\left( {2x + y + 2x - y} \right)\\
= 2y.4x = 8xy
\end{array}\)
