Question

Tính giá trị biểu thức:

\(A=\left(3x^3+8x^2+2\right)^{2005}\) biết \(x=\frac{\sqrt[3]{17\sqrt{5}-38}\left(\sqrt{5}+2\right)}{\sqrt{5}+\sqrt{14-6\sqrt{5}}}\)

Answer 1

\(x=\frac{\sqrt[3]{17\sqrt{5}-38}\left(\sqrt{5}+2\right)}{\sqrt{5}+\sqrt{\left(3-\sqrt{5}\right)^2}}=\frac{\sqrt[3]{17\sqrt{5}-38}\left(\sqrt{5}+2\right)}{3}=\frac{\sqrt[3]{17\sqrt{5}-38}.\sqrt[3]{\left(\sqrt{5}+2\right)^3}}{3}\)

\(=\frac{\sqrt[3]{\left(17\sqrt{5}-38\right)\left(17\sqrt{5}+38\right)}}{3}=\frac{1}{3}\)

\(\Rightarrow A=\left[3.\left(\frac{1}{3}\right)^3+8.\left(\frac{1}{3}\right)^2+2\right]^{2005}=3^{2005}\)