Đáp án: $x = - 3;x = - 1;x = 1$
Giải thích các bước giải:
$\begin{array}{l}
{\left( {{x^2} + 2x} \right)^2} - 2{x^2} - 4x = 3\\
\Leftrightarrow {\left( {{x^2} + 2x} \right)^2} - 2.\left( {{x^2} + 2x} \right) - 3 = 0\\
Đặt:\left( {{x^2} + 2x} \right) = a\\
\Leftrightarrow {a^2} - 2a - 3 = 0\\
\Leftrightarrow {a^2} - 3a + a - 3 = 0\\
\Leftrightarrow a.\left( {a - 3} \right) + \left( {a - 3} \right) = 0\\
\Leftrightarrow \left( {a - 3} \right)\left( {a + 1} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
a - 3 = 0\\
a + 1 = 0
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
{x^2} + 2x - 3 = 0\\
{x^2} + 2x + 1 = 0
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
{x^2} + 3x - x - 3 = 0\\
{\left( {x + 1} \right)^2} = 0
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
\left( {x + 3} \right)\left( {x - 1} \right) = 0\\
x + 1 = 0
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x + 3 = 0\\
x - 1 = 0\\
x + 1 = 0
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = - 3\\
x = 1\\
x = - 1
\end{array} \right.\\
Vậy\,x = - 3;x = - 1;x = 1
\end{array}$
