Đáp án đúng: A
Giải chi tiết:ĐK : \({{x}^{3}}-4\ge 0\Leftrightarrow x\ge \sqrt[3]{4}\).
\(\begin{array}{l}\,\,\,\,\,{\left( {{x^3} - 4} \right)^3} = {\left( {\sqrt[3]{{{{\left( {{x^2} + 4} \right)}^2}}} + 4} \right)^2}\\ \Leftrightarrow {\left( {{x^3} - 4} \right)^3} - {x^6} = {\left( {\sqrt[3]{{{{\left( {{x^2} + 4} \right)}^2}}} + 4} \right)^2} - {x^6}\\ \Leftrightarrow {\left( {{x^3} - 4} \right)^3} - {\left( {{x^2}} \right)^3} = {\left( {\sqrt[3]{{{{\left( {{x^2} + 4} \right)}^2}}} + 4} \right)^2} - {\left( {{x^2} + 4} \right)^2} + {\left( {{x^2} + 4} \right)^2} - {\left( {{x^3}} \right)^2}\\ \Leftrightarrow \left( {{x^3} - 4 - {x^2}} \right)\left[ {{{\left( {{x^3} - 4} \right)}^2} + {x^4} + {x^2}\left( {{x^3} - 4} \right)} \right]\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \left( {\sqrt[3]{{{{\left( {{x^2} + 4} \right)}^2}}} + 4 - {x^2} - 4} \right)\left( {\sqrt[3]{{{{\left( {{x^2} + 4} \right)}^2}}} + 4 + {x^2} + 4} \right) + \left( {{x^2} + 4 - {x^3}} \right)\left( {{x^2} + 4 + {x^3}} \right)\end{array}\)
\(\begin{array}{l} \Leftrightarrow \left( {{x^3} - 4 - {x^2}} \right)\left[ {{{\left( {{x^3} - 4} \right)}^2} + {x^4} + {x^2}\left( {{x^3} - 4} \right)} \right]\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \left( {\sqrt[3]{{{{\left( {{x^2} + 4} \right)}^2}}} - {x^2}} \right)\left( {\sqrt[3]{{{{\left( {{x^2} + 4} \right)}^2}}} + {x^2} + 8} \right) + \left( {{x^2} + 4 - {x^3}} \right)\left( {{x^2} + 4 + {x^3}} \right)\\ \Leftrightarrow \left( {{x^3} - 4 - {x^2}} \right)\left[ {{{\left( {{x^3} - 4} \right)}^2} + {x^4} + {x^2}\left( {{x^3} - 4} \right)} \right]\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{{{\left( {{x^2} + 4} \right)}^2} - {x^6}}}{{\left[ {\sqrt[3]{{{{\left( {{x^2} + 4} \right)}^4}}} + {x^4} + {x^2}\sqrt[3]{{{{\left( {{x^2} + 4} \right)}^2}}}} \right]}}\left( {\sqrt[3]{{{{\left( {{x^2} + 4} \right)}^2}}} + {x^2} + 8} \right) + \left( {{x^2} + 4 - {x^3}} \right)\left( {{x^2} + 4 + {x^3}} \right)\\ \Leftrightarrow \left( {{x^3} - 4 - {x^2}} \right)\left[ {{{\left( {{x^3} - 4} \right)}^2} + {x^4} + {x^2}\left( {{x^3} - 4} \right)} \right]\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{\left( {{x^2} + 4 - {x^3}} \right)\left( {{x^2} + 4 + {x^3}} \right)}}{{\left[ {\sqrt[3]{{{{\left( {{x^2} + 4} \right)}^4}}} + {x^4} + {x^2}\sqrt[3]{{{{\left( {{x^2} + 4} \right)}^2}}}} \right]}}\left( {\sqrt[3]{{{{\left( {{x^2} + 4} \right)}^2}}} + {x^2} + 8} \right) + \left( {{x^2} + 4 - {x^3}} \right)\left( {{x^2} + 4 + {x^3}} \right)\\ \Leftrightarrow \left( {{x^3} - 4 - {x^2}} \right)\left[ {{{\left( {{x^3} - 4} \right)}^2} + {x^4} + {x^2}\left( {{x^3} - 4} \right)} \right] - \frac{{\left( {{x^2} + 4 - {x^3}} \right)\left( {{x^2} + 4 + {x^3}} \right)}}{{\left[ {\sqrt[3]{{{{\left( {{x^2} + 4} \right)}^4}}} + {x^4} + {x^2}\sqrt[3]{{{{\left( {{x^2} + 4} \right)}^2}}}} \right]}}\left( {\sqrt[3]{{{{\left( {{x^2} + 4} \right)}^2}}} + {x^2} + 8} \right)\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, - \left( {{x^2} + 4 - {x^3}} \right)\left( {{x^2} + 4 + {x^3}} \right) = 0\\ \Leftrightarrow \left( {{x^3} - 4 - {x^2}} \right)\left\{ \begin{array}{l}\left[ {{{\left( {{x^3} - 4} \right)}^2} + {x^4} + {x^2}\left( {{x^3} - 4} \right)} \right]\\ + \frac{{\left( {{x^2} + 4 + {x^3}} \right)}}{{\left[ {\sqrt[3]{{{{\left( {{x^2} + 4} \right)}^4}}} + {x^4} + {x^2}\sqrt[3]{{{{\left( {{x^2} + 4} \right)}^2}}}} \right]}}\left( {\sqrt[3]{{{{\left( {{x^2} + 4} \right)}^2}}} + {x^2} + 8} \right) + \left( {{x^2} + 4 + {x^3}} \right)\end{array} \right\} = 0\,\,\,\,\left( 1 \right)\\Vi\,\,x \ge \sqrt[3]{4} > 0\\ \Leftrightarrow \left[ {{{\left( {{x^3} - 4} \right)}^2} + {x^4} + {x^2}\left( {{x^3} - 4} \right)} \right] + \frac{{\left( {{x^2} + 4 + {x^3}} \right)}}{{\left[ {\sqrt[3]{{{{\left( {{x^2} + 4} \right)}^4}}} + {x^4} + {x^2}\sqrt[3]{{{{\left( {{x^2} + 4} \right)}^2}}}} \right]}}\left( {\sqrt[3]{{{{\left( {{x^2} + 4} \right)}^2}}} + {x^2} + 8} \right) + \left( {{x^2} + 4 + {x^3}} \right) > 0\\ \Rightarrow \left( 1 \right) \Leftrightarrow {x^3} - 4 - {x^2} = 0 \Leftrightarrow {x^3} - 8 - {x^2} + 4 = 0\\ \Leftrightarrow \left( {x - 2} \right)\left( {{x^2} + 2x + 4} \right) - \left( {x - 2} \right)\left( {x + 2} \right) = 0\\ \Leftrightarrow \left( {x - 2} \right)\left( {{x^2} + 2x + 4 - x - 2} \right) = 0\\ \Leftrightarrow \left( {x - 2} \right)\left( {{x^2} + x + 2} \right) = 0 \Leftrightarrow x = 2\,\,\,\left( {tm} \right)\end{array}\)
Vậy nghiệm của phương trình là \(x=2\)
Chọn A.