Đáp án:
$\begin{array}{l}
2d)\\
{\left( {{x^2} - 4x + 6} \right)^2} - 4x\left( {{x^2} - 4x + 6} \right) + 3{x^2}\\
= {\left( {{x^2} - 4x + 6} \right)^2} - x.\left( {{x^2} - 4x + 6} \right)\\
- 3x\left( {{x^2} - 4x + 6} \right) + 3{x^2}\\
= \left( {{x^2} - 4x + 6} \right)\left( {{x^2} - 4x + 6 - x} \right)\\
- 3x.\left( {{x^2} - 4x + 6 - x} \right)\\
= \left( {{x^2} - 4x + 6} \right).\left( {{x^2} - 5x + 6} \right)\\
- 3x\left( {{x^2} - 5x + 6} \right)\\
= \left( {{x^2} - 5x + 6} \right)\left( {{x^2} - 4x + 6 - 3x} \right)\\
= \left( {{x^2} - 3x - 2x + 6} \right)\left( {{x^2} - 7x + 6} \right)\\
= \left( {x - 3} \right)\left( {x - 2} \right)\left( {x - 1} \right)\left( {x - 6} \right)\\
= \left( {x - 1} \right)\left( {x - 2} \right)\left( {x - 3} \right)\left( {x - 6} \right)\\
3c)\\
{\left( {{x^2} + 2x} \right)^2} - 2{x^2} - 4x = 3\\
\Rightarrow {\left( {{x^2} + 2x} \right)^2} - 2.\left( {{x^2} + 2x} \right) - 3 = 0\\
\Rightarrow {\left( {{x^2} + 2x} \right)^2} - 3\left( {{x^2} + 2x} \right) + \left( {{x^2} + 2x} \right) - 3 = 0\\
\Rightarrow \left( {{x^2} + 2x} \right)\left( {{x^2} + 2x - 3} \right) + \left( {{x^2} + 2x - 3} \right) = 0\\
\Rightarrow \left( {{x^2} + 2x - 3} \right)\left( {{x^2} + 2x + 1} \right) = 0\\
\Rightarrow \left( {x - 1} \right)\left( {x + 3} \right){\left( {x + 1} \right)^2} = 0\\
\Rightarrow \left[ \begin{array}{l}
x - 1 = 0\\
x + 3 = 0\\
x + 1 = 0
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
x = 1\\
x = - 3\\
x = - 1
\end{array} \right.\\
\text{Vậy}\,x = 1;x = - 1;x = - 3
\end{array}$