Đáp án:
$\begin{array}{l}
7){\left( {{x^2} + 4} \right)^2} - 16{x^2}\\
= \left( {{x^2} + 4 + 4x} \right)\left( {{x^2} + 4 - 4x} \right)\\
= {\left( {x + 2} \right)^2}{\left( {x - 2} \right)^2}\\
8){\left( {x + 1} \right)^3} - 8\\
= {\left( {x + 1} \right)^3} - {2^3}\\
= \left( {x + 1 - 2} \right)\left[ {{{\left( {x + 1} \right)}^2} + 2\left( {x + 1} \right) + 4} \right]\\
= \left( {x - 1} \right)\left( {{x^2} + 2x + 1 + 2x + 2 + 4} \right)\\
= \left( {x - 1} \right)\left( {{x^2} + 4x + 7} \right)\\
9){\left( {x + y} \right)^3} - {z^3}\\
= \left( {x + y - z} \right)\left[ {{{\left( {x + y} \right)}^2} + \left( {x + y} \right)z + {z^2}} \right]\\
= \left( {x + y - z} \right)\left( {{x^2} + {y^2} + {z^2} + 2xy + xz + yz} \right)\\
10)\\
10a.x - 5ay - 2x + y\\
= 5a.\left( {2x - y} \right) - \left( {2x - y} \right)\\
= \left( {2x - y} \right)\left( {5a - 1} \right)
\end{array}$
Câu 3 kiểm tra lại đề bài
