cho x²+1/x²=7.cmr: x⁵+1/x⁵ thuộc z Giúp mình vs mình cần gấpp ạ 17/07/2021 Bởi Maria cho x²+1/x²=7.cmr: x⁵+1/x⁵ thuộc z Giúp mình vs mình cần gấpp ạ
$\begin{array}{l} {x^2} + \dfrac{1}{{{x^2}}} = 7 \Leftrightarrow {\left( {x + \dfrac{1}{x}} \right)^2} – 2 = 7\\ \Leftrightarrow {\left( {x + \dfrac{1}{x}} \right)^2} = 9 \Leftrightarrow x + \dfrac{1}{x} = \pm 3 \in \mathbb{Z}\\ {x^3} + \dfrac{1}{{{x^3}}} = \left( {x + \dfrac{1}{x}} \right)\left( {{x^2} – 1 + \dfrac{1}{{{x^2}}}} \right) \in \mathbb{Z}\\ \Rightarrow {x^5} + \dfrac{1}{{{x^5}}} = \left( {{x^2} + \dfrac{1}{{{x^2}}}} \right)\left( {{x^3} + \dfrac{1}{{{x^3}}}} \right) – \left( {x + \dfrac{1}{x}} \right) \in \mathbb{Z} \end{array}$ Bình luận
$\begin{array}{l} {x^2} + \dfrac{1}{{{x^2}}} = 7 \Leftrightarrow {\left( {x + \dfrac{1}{x}} \right)^2} – 2 = 7\\ \Leftrightarrow {\left( {x + \dfrac{1}{x}} \right)^2} = 9 \Leftrightarrow x + \dfrac{1}{x} = \pm 3 \in \mathbb{Z}\\ {x^3} + \dfrac{1}{{{x^3}}} = \left( {x + \dfrac{1}{x}} \right)\left( {{x^2} – 1 + \dfrac{1}{{{x^2}}}} \right) \in \mathbb{Z}\\ \Rightarrow {x^5} + \dfrac{1}{{{x^5}}} = \left( {{x^2} + \dfrac{1}{{{x^2}}}} \right)\left( {{x^3} + \dfrac{1}{{{x^3}}}} \right) – \left( {x + \dfrac{1}{x}} \right) \in \mathbb{Z} \end{array}$