Đáp án:
\[P = 1\]
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
x = \dfrac{{\sqrt[3]{{10 + 6\sqrt 3 }}.\left( {\sqrt 3 - 1} \right)}}{{\sqrt {6 + 2\sqrt 5 } - \sqrt 5 }}\\
\Leftrightarrow x = \dfrac{{\sqrt[3]{{3\sqrt 3 + 9 + 3\sqrt 3 + 1}}.\left( {\sqrt 3 - 1} \right)}}{{\sqrt {5 + 2\sqrt 5 + 1} - \sqrt 5 }}\\
\Leftrightarrow x = \dfrac{{\sqrt[3]{{{{\left( {\sqrt 3 } \right)}^3} + 3.{{\left( {\sqrt[3]{3}} \right)}^2}.1 + 3.\sqrt 3 {{.1}^2} + {1^3}}}.\left( {\sqrt 3 - 1} \right)}}{{\sqrt {{{\left( {\sqrt 5 } \right)}^2} + 2.\sqrt 5 .1 + 1} - \sqrt 5 }}\\
\Leftrightarrow x = \dfrac{{\sqrt[3]{{{{\left( {\sqrt 3 + 1} \right)}^3}}}.\left( {\sqrt 3 - 1} \right)}}{{\sqrt {{{\left( {\sqrt 5 + 1} \right)}^2}} - \sqrt 5 }}\\
\Leftrightarrow x = \dfrac{{\left( {\sqrt 3 + 1} \right)\left( {\sqrt 3 - 1} \right)}}{{\sqrt 5 + 1 - \sqrt 5 }}\\
\Leftrightarrow x = \dfrac{{3 - 1}}{1}\\
\Leftrightarrow x = 2\\
\Rightarrow P = {\left( {{x^3} - 4x + 1} \right)^{2021}} = {\left( {{2^3} - 4.2 + 1} \right)^{2021}} = {1^{2021}} = 1
\end{array}\)
Vậy \(P = 1\)
