Đặt
$\begin{array}{l}
a = \sqrt[3]{{\dfrac{{{x^3} - 3x + \left( {{x^2} - 1} \right)\sqrt {{x^2} - 4} }}{2}}}\\
b = \sqrt[3]{{\dfrac{{{x^3} - 3x - \left( {{x^2} - 1} \right)\sqrt {{x^2} - 4} }}{2}}}\\
\Rightarrow A = a + b\\
\Rightarrow ab = \sqrt[3]{{\dfrac{{{{\left( {{x^3} - 3x} \right)}^2} - \left( {{x^4} - 2{x^2} + 1} \right)\left( {{x^2} - 4} \right)}}{4}}}\\
\Rightarrow ab = \sqrt[3]{{\dfrac{{{x^6} - 6{x^4} + 9{x^2} - \left( {{x^6} - 4{x^4} - 2{x^4} + 8{x^2} + {x^2} - 4} \right)}}{4}}}\\
\Rightarrow ab = \sqrt[3]{{\dfrac{4}{4}}} = 1\\
\Rightarrow {A^3} = {\left( {a + b} \right)^3}\\
= {a^3} + {b^3} + 3ab\left( {a + b} \right)\\
= \dfrac{{{x^3} - 3x + \left( {{x^2} - 1} \right)\sqrt {{x^2} - 4} + {x^3} - 3x - \left( {{x^2} - 1} \right)\sqrt {{x^2} - 4} }}{2} + 3ab\left( {a + b} \right)\\
= \dfrac{{2{x^3} - 6x}}{2} + 3A\\
\Rightarrow {A^3} - 3A - {x^3} + 3x = 0\\
\Rightarrow \left( {A - x} \right)\left( {{A^2} + Ax + {x^2}} \right) - 3\left( {A - x} \right) = 0\\
\Leftrightarrow \left( {A - x} \right)\left( {{A^2} + Ax + {x^2} - 3} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
x = A\\
{A^2} + Ax + {x^2} - 3 = 0\left( 1 \right)
\end{array} \right.\\
\left( 1 \right):A > 0,x = \sqrt[3]{{1995}},{x^2} = \sqrt[3]{{{{1995}^2}}}\\
\Rightarrow {A^2} + Ax + {x^2} - 3 > 3\\
\Rightarrow A = x = \sqrt[3]{{1995}}\\
\end{array}$
Vậy$A=\sqrt[3]{1995}$
