Giải thích các bước giải:

$\begin{array}{l}
B1:\\
a){x^2} - 3x + xy - 3y\\
 = \left( {{x^2} + xy} \right) - \left( {3x + 3y} \right)\\
 = \left( {x + y} \right)\left( {x - 3} \right)\\
b){x^2} + {y^2} - 2xy - 25\\
 = \left( {{x^2} - 2xy + {y^2}} \right) - 25\\
 = {\left( {x - y} \right)^2} - 25\\
 = \left( {x - y - 5} \right)\left( {x - y + 5} \right)
\end{array}$

$B2:$

Ta có:

$3{x^4} + 4x - 2{x^3} - 2{x^2} - 8 = 3{x^4} - 2{x^3} - 2{x^2} + 4x - 8$

Khi đó:

$\begin{array}{l}
3{x^4} - 2{x^3} - 2{x^2} + 4x - 8\\
 = 3{x^2}\left( {{x^2} - 2} \right) - 2x\left( {{x^2} - 2} \right) + 4{x^2} - 8\\
 = 3{x^2}\left( {{x^2} - 2} \right) - 2x\left( {{x^2} - 2} \right) + 4\left( {{x^2} - 2} \right)\\
 = \left( {{x^2} - 2} \right)\left( {3{x^2} - 2x + 4} \right)\\
 \Rightarrow \left( {3{x^4} - 2{x^3} - 2{x^2} + 4x - 8} \right):\left( {{x^2} - 2} \right) = 3{x^2} - 2x + 4
\end{array}$

Vậy $\left( {3{x^4} - 2{x^3} - 2{x^2} + 4x - 8} \right):\left( {{x^2} - 2} \right) = 3{x^2} - 2x + 4$

$\begin{array}{l}
B3:\\
a)\left( {x + 3} \right)\left( {{x^2} - 3x + 9} \right) - x{\left( {x - 2} \right)^2} = 27\\
 \Leftrightarrow \left( {{x^3} + {3^3}} \right) - x\left( {{x^2} - 4x + 4} \right) = 27\\
 \Leftrightarrow 4{x^2} - 4x = 0\\
 \Leftrightarrow 4x\left( {x - 1} \right) = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
x = 0\\
x - 1 = 0
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
x = 0\\
x = 1
\end{array} \right.
\end{array}$

Vậy $x \in \left\{ {0;1} \right\}$

$\begin{array}{l}
b)\left( {x - 1} \right)\left( {x - 5} \right) + 3 = 0\\
 \Leftrightarrow {x^2} - 6x + 5 + 3 = 0\\
 \Leftrightarrow {x^2} - 6x + 8 = 0\\
 \Leftrightarrow \left( {x - 2} \right)\left( {x - 4} \right) = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
x - 2 = 0\\
x - 4 = 0
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
x = 2\\
x = 4
\end{array} \right.
\end{array}$

Vậy $x \in \left\{ {2;4} \right\}$