Đáp án:
$\begin{array}{l}
1)16{x^3}y + 0,25y{z^3}\\
= \frac{1}{4}y\left( {64{x^3} + {z^3}} \right)\\
= \frac{1}{4}y\left( {4x + z} \right)\left( {16{x^2} - 4xz + {z^2}} \right)\\
2){x^4} - 4{x^3} + 4{x^2}\\
= {x^2}\left( {{x^2} - 4x + 4} \right)\\
= {x^2}{\left( {x - 2} \right)^2}\\
3)2a{b^2} - {a^2}b - {b^3}\\
= - b\left( { - 2ab + {a^2} + {b^2}} \right)\\
= - b{\left( {a - b} \right)^2}\\
4){a^3} + {a^2}b - a{b^2} - {b^3}\\
= {a^2}\left( {a + b} \right) - {b^2}\left( {a + b} \right)\\
= \left( {a + b} \right)\left( {{a^2} - {b^2}} \right)\\
= {\left( {a + b} \right)^2}\left( {a - b} \right)\\
5){x^3} + {x^2} - 4x - 4\\
= \left( {{x^2} - 4} \right)\left( {x + 1} \right)\\
= \left( {x - 2} \right)\left( {x + 2} \right)\left( {x + 1} \right)\\
6){x^3} - {x^2} - x + 1\\
= \left( {x - 1} \right)\left( {{x^2} - 1} \right)\\
= {\left( {x - 1} \right)^2}\left( {x + 1} \right)\\
7){x^4} + {x^3} + {x^2} - 1\\
= \left( {x + 1} \right)\left( {{x^3} + x - 1} \right)\\
8){x^2}{y^2} + 1 - {x^2} - {y^2}\\
= {x^2}\left( {{y^2} - 1} \right) - \left( {{y^2} - 1} \right)\\
= \left( {{y^2} - 1} \right)\left( {{x^2} - 1} \right)\\
= \left( {y - 1} \right)\left( {y + 1} \right)\left( {x - 1} \right)\left( {x + 1} \right)\\
10){x^4} - {x^2} + 2x - 1\\
= {x^4} - \left( {{x^2} - 2x + 1} \right)\\
= {\left( {{x^2}} \right)^2} - {\left( {x - 1} \right)^2}\\
= \left( {{x^2} - x + 1} \right)\left( {{x^2} + x - 1} \right)\\
11)3a - 3b + {a^2} - 2ab + {b^2}\\
= 3\left( {a - b} \right) + {\left( {a - b} \right)^2}\\
= \left( {a - b} \right)\left( {3 + a + b} \right)\\
12){a^2} + 2ab + {b^2} - 2a - 2b + 1\\
= {\left( {a + b} \right)^2} - 2\left( {a + b} \right) + 1\\
= {\left( {a + b - 1} \right)^2}\\
13){a^2} - {b^2} - 4a + 4b\\
= \left( {a - b} \right)\left( {a + b - 4} \right)\\
14){a^3} - {b^3} - 3a + 3b\\
= \left( {a - b} \right)\left( {{a^2} + ab + {b^2} - 3} \right)\\
15){x^3} + 3{x^2} - 3x - 1\\
= {x^3} - 1 + 3x\left( {x - 1} \right)\\
= \left( {x - 1} \right)\left( {{x^2} + x + 1 + 3x} \right)\\
= \left( {x - 1} \right)\left( {{x^2} + 4x + 1} \right)\\
16){x^3} - 3{x^2} - 3x + 1\\
= {x^3} + 1 - 3x\left( {x + 1} \right)\\
= \left( {x + 1} \right)\left( {{x^2} - x + 1} \right) - 3x\left( {x + 1} \right)\\
= \left( {x + 1} \right)\left( {{x^2} - x + 1 - 3x} \right)\\
= \left( {x + 1} \right)\left( {{x^2} - 4x + 1} \right)\\
17){x^3} - 4{x^2} + 4x - 1\\
= {x^3} - 1 - 4{x^2} + 4x\\
= \left( {x - 1} \right)\left( {{x^2} + x + 1} \right) - 4x\left( {x - 1} \right)\\
= \left( {x - 1} \right)\left( {{x^2} + x + 1 - 4x} \right)\\
= \left( {x - 1} \right)\left( {{x^2} - 3x + 1} \right)
\end{array}$
