Đáp án:
\[A = 2021\]
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
x = \sqrt 5 + 1\\
\Leftrightarrow x - 1 = \sqrt 5 \\
\Leftrightarrow {\left( {x - 1} \right)^2} = 5\\
\Leftrightarrow {x^2} - 2x + 1 = 5\\
\Leftrightarrow {x^2} - 2x - 4 = 0\\
A = \dfrac{{{x^5} - 2{x^4} - 3{x^3} - 3{x^2} - 2x - 6059}}{{{x^2} - 2x - 7}}\\
= \dfrac{{\left( {{x^5} - 2{x^4} - 4{x^3}} \right) + \left( {{x^3} - 2{x^2} - 4x} \right) - \left( {{x^2} - 2x - 4} \right) - 6063}}{{\left( {{x^2} - 2x - 4} \right) - 3}}\\
= \dfrac{{{x^3}.\left( {{x^2} - 2x - 4} \right) + x.\left( {{x^2} - 2x - 4} \right) - \left( {{x^2} - 2x - 4} \right) - 6063}}{{\left( {{x^2} - 2x - 4} \right) - 3}}\\
= \dfrac{{{x^3}.0 + x.0 - 0 - 6063}}{{0 - 3}}\,\,\,\,\,\,\,\left( {{x^2} - 2x - 4 = 0} \right)\\
= \dfrac{{6063}}{3} = 2021
\end{array}\)
Vậy \(A = 2021\)
