Đáp án:
$\begin{array}{l}
a){\left( {x + 1} \right)^2} = 4{\left( {{x^2} - 2x + 1} \right)^2}\\
\Rightarrow {\left( {x + 1} \right)^2} = {\left( {2{x^2} - 4x + 2} \right)^2}\\
\Rightarrow \left[ \begin{array}{l}
x + 1 = 2{x^2} - 4x + 2\\
x + 1 = - 2{x^2} + 4x - 2
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
2{x^2} - 4x + 1 = 0\\
2{x^2} + 3x + 3 = 0\left( {vn} \right)
\end{array} \right.\\
\Rightarrow x = \dfrac{{2 \pm \sqrt 2 }}{2}\\
b){\left( {{x^2} - 9} \right)^2} - 9{\left( {x - 3} \right)^2} = 0\\
\Rightarrow {\left( {x - 3} \right)^2}{\left( {x + 3} \right)^2} - 9{\left( {x - 3} \right)^2} = 0\\
\Rightarrow \left[ \begin{array}{l}
{\left( {x - 3} \right)^2} = 0\\
{\left( {x + 3} \right)^2} = 9
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
x = 3\\
x + 3 = 3\\
x + 3 = - 3
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
x = 3\\
x = 0\\
x = - 6
\end{array} \right.\\
Vay\,x \in \left\{ { - 6;0;3} \right\}\\
c)9{\left( {x - 3} \right)^2} = 4{\left( {x + 2} \right)^2}\\
\Rightarrow {\left( {3x - 9} \right)^2} = {\left( {2x + 4} \right)^2}\\
\Rightarrow \left[ \begin{array}{l}
3x - 9 = 2x + 4\\
3x - 9 = - 2x - 4
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
x = 13\\
x = 1
\end{array} \right.\\
d){\left( {4{x^2} - 3x - 18} \right)^2} = {\left( {4{x^2} + 3x} \right)^2}\\
\Rightarrow \left[ \begin{array}{l}
4{x^2} - 3x - 18 = 4{x^2} + 3x\\
4{x^2} - 3x - 18 = - 4{x^2} - 3x
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
6x = - 18\\
- 18 = 0\left( {ktm} \right)
\end{array} \right.\\
\Rightarrow x = - 3\\
Vay\,x = - 3
\end{array}$
