Question

[Hạ Vy]

tìm 1 đa thức có hệ số nguyên bậc 7 nhận x=\(\sqrt[7]{\frac{2}{5}}+\sqrt[7]{\frac{5}{2}}\) là nghiệm

Answer 1

[Nguyễn Việt Lâm]

Đặt \(a=\sqrt[7]{\frac{2}{5}}\Rightarrow x=a+\frac{1}{a}\Rightarrow\left\{{}\begin{matrix}x^2=a^2+\frac{1}{a^2}+2\\x^3=a^3+\frac{1}{a^3}+3x\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x^2-2=a^2+\frac{1}{a^2}\\x^3-3x=a^3+\frac{1}{a^3}\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x^4-4x^2+4=a^4+\frac{1}{a^4}+2\\x^3-3x=a^3+\frac{1}{a^3}\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x^4-4x^2+2=a^4+\frac{1}{a^4}\\x^3-3x=a^3+\frac{1}{a^3}\end{matrix}\right.\)

\(\Rightarrow\left(x^4-4x^2+2\right)\left(x^3-3x\right)=\left(a^4+\frac{1}{a^4}\right)\left(a^3+\frac{1}{a^3}\right)\)

\(\Leftrightarrow x^7-7x^5+14x^3-6x=a^7+\frac{1}{a^7}+a+\frac{1}{a}\)

\(\Leftrightarrow x^7-7x^5+14x^3-6x=\frac{2}{5}+\frac{5}{2}+x\)

\(\Leftrightarrow x^7-7x^5+14x^3-7x-\frac{29}{10}=0\)

\(\Leftrightarrow10x^7-70x^5+140x^3-70x-29=0\)

Đây là 1 trong những pt có hệ số nguyên cần tìm