已知：\(x^2+\frac{1}{x^2} = \sqrt{2}\)，
求解：\(x^{2022}+\frac{1}{x^{2022}}\)
已知：\(x^2+\frac{1}{x^2} = \sqrt{2}\)
\(x^4+\frac{1}{x^4}+2=2\)
\(x^4+\frac{1}{x^4}=0\)
\(x^8+1=0\)
得：\(x^8=-1\)
解：\(x^{2022}+\frac{1}{x^{2022}}\)
=\(x^{2024-2}+\frac{1}{x^{2024-2}}\)
=\(\frac{x^{2024}}{x^2}+\frac{1}{\frac{x^{2024}}{x^2}}\)
=\(\frac{{x^8}^{253}}{x^2}+\frac{x^2}{{x^8}^{253}}\)
=\(\frac{{-1}^{253}}{x^2}+\frac{x^2}{{-1}^{253}}\)
=\(\frac{-1}{x^2}+\frac{x^2}{-1}\)
=\(-1(x^2+\frac{1}{x^2})\)
=\(-\sqrt{2}\)