Question

Giải hệ phương trình: \(\left\{{}\begin{matrix}x^3-3x=y^3-3y\\x^6+y^6=1\end{matrix}\right.\)

Answer 1

@Nguyễn Việt Lâm

Answer 2

\(1=x^6+y^6=\left(x^2\right)^3+\left(y^2\right)^3\ge\frac{1}{4}\left(x^2+y^2\right)^3\)

\(\Rightarrow x^2+y^2\le\sqrt[3]{4}\)

\(\Rightarrow2xy\le\sqrt[3]{4}\Rightarrow xy\le\frac{\sqrt[3]{4}}{2}\)

\(\Rightarrow x^2+xy+y^2\le\frac{3\sqrt[3]{4}}{2}< \frac{3\sqrt[3]{8}}{2}=3\)

Xét pt đầu:

\(x^3-y^3-3\left(x-y\right)=0\)

\(\Leftrightarrow\left(x-y\right)\left(x^2+xy+y^2\right)-3\left(x-y\right)=0\)

\(\Leftrightarrow\left(x-y\right)\left(x^2+xy+y^2-3\right)=0\)

\(\Leftrightarrow x=y\) (do \(x^2+xy+y^2-3< 0\) theo cmt)

Thế xuống pt dưới:

\(2x^6=1\Leftrightarrow x=y=\pm\frac{1}{\sqrt[6]{2}}\)