Điều kiện xác định $x\ge-2$
$\begin{array}{l} \left( {{x^4} - 11{x^2} + 18} \right)\sqrt {x + 2} = 0\\ \Leftrightarrow \left[ \begin{array}{l} {x^4} - 11{x^2} + 18 = 0\\ \sqrt {x + 2} = 0 \end{array} \right. \Leftrightarrow \left[ \begin{array}{l} x = - 2\\ \left( {{x^2} - 9} \right)\left( {{x^2} - 2} \right) = 0 \end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l} x = - 2\\ {x^2} = 9\\ {x^2} = 2 \end{array} \right. \Leftrightarrow \left[ \begin{array}{l} x = - 2\\ x = 3\\ x = - 3\\ x = \sqrt 2 \\ x = - \sqrt 2 \end{array} \right. \Rightarrow \left[ \begin{array}{l} x = - 2\\ x = 3\\ x = \sqrt 2 \\ x = - \sqrt 2 \end{array} \right.\left( {x \ge - 2} \right) \end{array}$
Chọn D
