Đáp án:

$\begin{array}{l}
1)a)3{x^2} - 6x = 3x\left( {x - 2} \right)\\
b){x^2} - 2x + 1 - {y^2}\\
 = {\left( {x - 1} \right)^2} - {y^2}\\
 = \left( {x - 1 - y} \right)\left( {x - 1 + y} \right)\\
c)9{x^3} - 9{x^2}y - 4x + 4y\\
 = 9{x^2}\left( {x - y} \right) - 4\left( {x - y} \right)\\
 = \left( {x - y} \right)\left( {9{x^2} - 4} \right)\\
 = \left( {x - y} \right)\left( {3x - 2} \right)\left( {3x + 2} \right)\\
d){x^3} - 2{x^2} - 8x\\
 = x\left( {{x^2} - 2x - 8} \right)\\
 = x\left( {{x^2} - 4x + 2x - 8} \right)\\
 = x\left( {x - 4} \right)\left( {x + 2} \right)\\
e)4{x^2} - 12xy + 5{y^2}\\
 = 4{x^2} - 10xy - 2xy + 5{y^2}\\
 = 2x\left( {2x - 5y} \right) - y\left( {2x - 5y} \right)\\
 = \left( {2x - 5y} \right)\left( {2x - y} \right)\\
f){\left( {x + y + 2z} \right)^2} + {\left( {x + y - z} \right)^2} - 9{z^2}\\
 = {\left( {x + y + 2z} \right)^2} + \left( {x + y - z - 3z} \right)\left( {x + y - z + 3z} \right)\\
 = {\left( {x + y + 2z} \right)^2} + \left( {x + y - 4z} \right)\left( {x + y + 2z} \right)\\
 = \left( {x + y + 2z} \right)\left( {x + y + 2z + x + y - 4z} \right)\\
 = \left( {x + y + 2z} \right)\left( {2x + 2y - 2z} \right)\\
 = 2\left( {x + y + 2z} \right)\left( {x + y - z} \right)\\
g){x^4} + 2019{x^2} + 2018x + 2019\\
 = {x^4} - x + 2019{x^2} + 2019x + 2019\\
 = x\left( {{x^3} - 1} \right) + 2019\left( {{x^2} + x + 1} \right)\\
 = x\left( {x - 1} \right)\left( {{x^2} + x + 1} \right) + 2019\left( {{x^2} + x + 1} \right)\\
 = \left( {{x^2} + x + 1} \right)\left( {x\left( {x - 1} \right) + 2019} \right)\\
 = \left( {{x^2} + x + 1} \right)\left( {{x^2} - x + 2019} \right)\\
B2)\\
a)x\left( {x - 1} \right) - {x^2} + 2x = 5\\
 \Leftrightarrow {x^2} - x - {x^2} + 2x = 5\\
 \Leftrightarrow x = 5\\
Vậy\,x = 5\\
b)4{x^3} - 36x = 0\\
 \Leftrightarrow 4x\left( {x - 9} \right) = 0\\
 \Leftrightarrow x = 0;x = 9\\
Vậy\,x = 0;x = 9\\
c)2{x^2} - 2x = {\left( {x - 1} \right)^2}\\
 \Leftrightarrow 2x\left( {x - 1} \right) - {\left( {x - 1} \right)^2} = 0\\
 \Leftrightarrow \left( {x - 1} \right)\left( {2x - x + 1} \right) = 0\\
 \Leftrightarrow \left( {x - 1} \right)\left( {x + 1} \right) = 0\\
 \Leftrightarrow x = 1;x =  - 1\\
Vậy\,x = 1;x =  - 1\\
d)\left( {x - 7} \right)\left( {{x^2} - 9x + 20} \right)\left( {x - 2} \right) = 72\\
 \Leftrightarrow \left( {x - 7} \right)\left( {x - 2} \right)\left( {{x^2} - 9x + 20} \right) = 72\\
 \Leftrightarrow \left( {{x^2} - 9x + 14} \right)\left( {{x^2} - 9x + 20} \right) = 72\\
Dat:\left( {{x^2} - 9x + 14} \right) = a\\
 \Leftrightarrow a\left( {a + 6} \right) = 72\\
 \Leftrightarrow {a^2} + 6a - 72 = 0\\
 \Leftrightarrow \left( {a + 12} \right)\left( {a - 6} \right) = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
a + 12 = 0\\
a - 6 = 0
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
{x^2} - 9x + 14 + 12 = 0\\
{x^2} - 9x + 14 - 6 = 0
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
{x^2} - 9x + 26 = 0\left( {vn} \right)\\
{x^2} - 9x + 8 = 0
\end{array} \right.\\
 \Leftrightarrow \left( {x - 1} \right)\left( {x - 8} \right) = 0\\
 \Leftrightarrow x = 1;x = 8\\
Vậy\,x = 1;x = 8
\end{array}$