Đáp án:
$\begin{array}{l}
1){x^2} - 6x + 9 = {\left( {x - 3} \right)^2}\\
2)25 + 10x + {x^2} = {\left( {5 + x} \right)^2}\\
3)\dfrac{1}{4}{a^2} + 2a{b^2} + 4{b^4}\\
= {\left( {\dfrac{1}{2}a + 2{b^2}} \right)^2}\\
4)\dfrac{1}{9} - \dfrac{2}{3}{y^4} + {y^8} = {\left( {\dfrac{1}{3} - {y^4}} \right)^2}\\
5){x^3} + 8{y^3} = \left( {x + 2y} \right)\left( {{x^2} - 2xy + 4{y^2}} \right)\\
6)8{y^3} - 125 = \left( {2y - 5} \right)\left( {4{y^2} + 10y + 25} \right)\\
7){a^6} - {b^3} = \left( {{a^2} - b} \right)\left( {{a^4} + {a^2}b + {b^2}} \right)\\
8){x^2} - 10x + 25 = {\left( {x - 5} \right)^2}\\
9)8{x^3} - \dfrac{1}{8} = \left( {2x - \dfrac{1}{2}} \right)\left( {4{x^2} + x + \dfrac{1}{4}} \right)\\
10){x^2} + 4xy + 4{y^2} = {\left( {x + 2y} \right)^2}\\
8){\left( {3x + 2} \right)^2} - 4\\
= \left( {3x + 2 - 2} \right)\left( {3x + 2 + 2} \right)\\
= 3x\left( {3x + 4} \right)\\
9)4{x^2} - 25{y^2} = \left( {2x - 5y} \right)\left( {4{x^2} - 10xy + 25{y^2}} \right)\\
10)4{x^2} - 49 = \left( {2x - 7} \right)\left( {2x + 7} \right)\\
11)8{z^3} + 27 = \left( {2z + 3} \right)\left( {4{z^2} - 6z + 9} \right)\\
12)\dfrac{9}{{25}}{x^4} - \dfrac{1}{4} = \left( {\dfrac{9}{5}{x^2} - \dfrac{1}{2}} \right)\left( {\dfrac{9}{5}{x^2} + \dfrac{1}{2}} \right)\\
13){x^{32}} - 1 = \left( {{x^{16}} - 1} \right)\left( {{x^{16}} + 1} \right)\\
= \left( {{x^8} - 1} \right).\left( {{x^8} + 1} \right)\left( {{x^{16}} + 1} \right)\\
= \left( {{x^4} - 1} \right)\left( {{x^4} + 1} \right)\left( {{x^8} + 1} \right)\left( {{x^{16}} + 1} \right)\\
= \left( {{x^2} - 1} \right).\left( {{x^2} + 1} \right).\left( {{x^4} + 1} \right)\left( {{x^8} + 1} \right)\left( {{x^{16}} + 1} \right)\\
= \left( {x - 1} \right)\left( {x + 1} \right)\left( {{x^2} + 1} \right).\left( {{x^4} + 1} \right)\left( {{x^8} + 1} \right)\left( {{x^{16}} + 1} \right)\\
14)4{x^2} + 4x + 1 = {\left( {2x + 1} \right)^2}\\
15){x^2} - 20x + 100 = {\left( {x - 10} \right)^2}\\
16){y^4} - 14{y^2} + 49 = {\left( {{y^2} - 7} \right)^2}\\
17)125{x^3} - 64{y^3}\\
= \left( {5x - 4y} \right)\left( {25{x^2} - 20xy + 16{y^2}} \right)
\end{array}$
