Question

cho x >0 thõa mãn \(x^2\)+ \(\dfrac{1}{x^2}\)= 7. Cmr:

\(x^5\)+\(\dfrac{1}{x^5}\) là một số nguyên. Tìm số nguyên đó.

Answer 1

\(x^2+\dfrac{1}{x^2}=7\Rightarrow\left(x+\dfrac{1}{x}\right)^2-2=7\Rightarrow x+\dfrac{1}{x}=3\)

\(\Rightarrow\left(x+\dfrac{1}{x}\right)\left(x^2+\dfrac{1}{x^2}\right)=21\)

\(\Rightarrow x^3+\dfrac{1}{x^3}+x+\dfrac{1}{x}=21\Rightarrow x^3+\dfrac{1}{x^3}=18\)

\(\Rightarrow\left(x^3+\dfrac{1}{x^3}\right)\left(x^2+\dfrac{1}{x^2}\right)=126\)

\(\Rightarrow x^5+\dfrac{1}{x^5}+x+\dfrac{1}{x}=126\Rightarrow x^5+\dfrac{1}{x^5}=123\)