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[Khoa Linh]

Cho $x=\frac{1}{3}\left ( 1-\sqrt[3]{\frac{25+\sqrt{621}}{2}}-\sqrt[3]{\frac{25-\sqrt{621}}{2}} \right )$

Chứng minh rằng $x^5+x+1=0$

[Khoa Linh]

Cho $x=\frac{1}{3}\left ( 1-\sqrt[3]{\frac{25+\sqrt{621}}{2}}-\sqrt[3]{\frac{25-\sqrt{621}}{2}} \right )$

Chứng minh rằng $x^5+x+1=0$

$GT\Leftrightarrow 1-3x=\sqrt[3]{\frac{25+\sqrt{621}}{2}}+\sqrt[3]{\frac{25-\sqrt{621}}{2}}$

$\Leftrightarrow (1-3x)^3=25+3(1-3x) \Leftrightarrow x^3-x^2+1=0\Leftrightarrow (x^2+x+1)(x^3-x^2+1)=0\Leftrightarrow x^5+x+1=0(q.e.d)$

Nguồn: Đỗ Xuân Trọng 

Edited by Khoa Linh, 03-02-2018 - 13:17.