Đáp án:
Bạn tham khảo nhé!!!
Giải thích các bước giải:
$\begin{array}{l}
1)\,\,\,16{x^3} + 54{y^3} = 2\left( {8{x^3} + 27{y^3}} \right)\\
= 2\left( {2x + 3y} \right)\left( {4{x^2} - 6xy + 9{y^2}} \right).\\
2)\,\,{x^2} - 2xy + {y^2} - 16 = {\left( {x - y} \right)^2} - 16\\
= \left( {x - y - 4} \right)\left( {x - y + 4} \right).\\
3)\,\,{x^3} + {x^2} - x + 2 = {x^3} + 2{x^2} - {x^2} - 2x + x + 2\\
= {x^2}\left( {x + 2} \right) - x\left( {x + 2} \right) + \left( {x + 2} \right)\\
= \left( {x + 2} \right)\left( {{x^2} - x + 1} \right).\\
4)\,\,2{x^2} - 2{y^2} - 6x - 6y = 2\left( {{x^2} - {y^2}} \right) - 6\left( {x + y} \right)\\
= 2\left( {x - y} \right)\left( {x + y} \right) - 6\left( {x + y} \right)\\
= 2\left( {x + y} \right)\left( {x - y - 3} \right).\\
5)\,\,\,{x^4} - 5{x^2} + 4 = {x^4} - {x^2} - 4{x^2} + 4\\
= {x^2}\left( {{x^2} - 1} \right) - 4\left( {{x^2} - 1} \right)\\
= \left( {{x^2} - 1} \right)\left( {{x^2} - 4} \right)\\
= \left( {x - 1} \right)\left( {x + 1} \right)\left( {x - 2} \right)\left( {x + 2} \right).
\end{array}$
