Đáp án:
$\begin{array}{l}
+ {x^2} - {y^2} - x + y\\
= \left( {x - y} \right)\left( {x + y} \right) - \left( {x - y} \right)\\
= \left( {x - y} \right)\left( {x + y - 1} \right)\\
+ 3{x^2}\left( {a + b + c} \right) + 36xy\left( {a + b + c} \right) + 108{y^2}\left( {a + b + c} \right)\\
= 3\left( {a + b + c} \right)\left( {{x^2} + 12xy + 36{y^2}} \right)\\
= 3\left( {a + b + c} \right){\left( {x + 6y} \right)^2}\\
+ ){x^4} + 4{x^2} - 5\\
= {x^4} + 5{x^2} - {x^2} - 5\\
= \left( {{x^2} + 5} \right)\left( {{x^2} - 1} \right)\\
= \left( {{x^2} + 5} \right)\left( {x + 1} \right)\left( {x - 1} \right)\\
+ ){x^3} - 19x - 30\\
= {x^3} + 2{x^2} - 2{x^2} - 4x - 15x - 30\\
= \left( {x + 2} \right)\left( {{x^2} - 2x - 15} \right)\\
= \left( {x + 2} \right)\left( {x + 3} \right)\left( {x - 5} \right)\\
+ ){x^4} + {x^2} + 1\\
= {x^4} + 2{x^2} + 1 - {x^2}\\
= {\left( {{x^2} + 1} \right)^2} - {x^2}\\
= \left( {{x^2} + x + 1} \right)\left( {{x^2} - x + 1} \right)\\
+ )2{x^3} - 5{x^2} + 8x - 3\\
= 2{x^3} - {x^2} - 4{x^2} + 2x + 6x - 3\\
= \left( {2x - 1} \right)\left( {{x^2} - 2x + 3} \right)\\
+ )3{x^3} - 14{x^2} + 4x + 3\\
= 3{x^3} + {x^2} - 15{x^2} - 5x + 9x + 3\\
= \left( {3x + 1} \right)\left( {{x^2} - 5x + 3} \right)
\end{array}$
$\begin{array}{l}
12{x^2} + 5x - 12{y^2} + 12y - 10xy - 3\\
= 12{x^2} - 18xy + 9x\\
- \left( {4x - 6y + 3} \right) + \left( {8xy - 12{y^2} + 6y} \right)\\
= 3x.\left( {4x - 6y + 3} \right) - \left( {4x - 6y + 3} \right)\\
+ 2y\left( {4x - 6y + 3} \right)\\
= \left( {4x - 6y + 3} \right)\left( {3x + 2y - 1} \right)
\end{array}$
