Đáp án:
$\begin{array}{l}
1a)4a - 6b = 2.\left( {2a - 3b} \right)\\
b)6xy - 3{x^2}y = 3xy\left( {2 - x} \right)\\
c)5{a^2}b - 10a{b^2} = 5ab\left( {a - 2b} \right)\\
d){x^2} + 6xy + 9{y^2}\\
= {\left( {x + 3y} \right)^2}\\
e){a^2} + 4a + 4 = {\left( {a + 2} \right)^2}\\
f)25 - 10a + {a^2} = {\left( {5 - a} \right)^2}\\
g){x^2} - 4x + 4 - {b^2}\\
= {\left( {x - 2} \right)^2} - {b^2}\\
= \left( {x - 2 + b} \right)\left( {x - 2 - b} \right)\\
h){a^2} - 4 + 4b - {b^2}\\
= {a^2} - \left( {{b^2} - 4b + 4} \right)\\
= {a^2} - {\left( {b - 2} \right)^2}\\
= \left( {a - b + 2} \right)\left( {a + b - 2} \right)\\
2)\\
a){x^2} - 8x = 0\\
\Leftrightarrow x\left( {x - 8} \right) = 0\\
\Leftrightarrow x = 0;x = 8\\
Vậy\,x = 0;x = 8\\
b){x^2} - 4x + 4 = 0\\
\Leftrightarrow {\left( {x - 2} \right)^2} = 0\\
\Leftrightarrow x - 2 = 0\\
\Leftrightarrow x = 2\\
Vậy\,x = 2
\end{array}$
