Cho x=$\sqrt{5}$ + 1 A= $\frac{x^{5}-2x^{4}-3x^{3}-3x^{2}-2x-6059}{x^{2}-2x-7}$ 18/08/2021 Bởi Hailey Cho x=$\sqrt{5}$ + 1 A= $\frac{x^{5}-2x^{4}-3x^{3}-3x^{2}-2x-6059}{x^{2}-2x-7}$
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 \(A = 2021\) Bình luận
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 \(A = 2021\)