Question

\(\left\{{}\begin{matrix}x^2+y^2=2xy+1\\x^5+y^3+1=0\end{matrix}\right.\)

Answer 1

\(x^2+y^2-2xy=1\Leftrightarrow\left(x-y\right)^2=1\Rightarrow\left[{}\begin{matrix}x-y=1\\x-y=-1\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}y=x+1\\y=x-1\end{matrix}\right.\)

Thay xuống dưới: \(\left[{}\begin{matrix}x^5+\left(x+1\right)^3+1=0\\x^5+\left(x-1\right)^3+1=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\left(x+1\right)\left(x^4-x^3+2x^2+x+2\right)=0\\x\left(x^4+x^2-3x+3\right)=0\end{matrix}\right.\)

Dễ dàng thấy: \(x^4-x^3+2x^2+x+2=\left(x^2-\frac{x}{2}\right)^2+\frac{7}{4}\left(x+\frac{2}{7}\right)^2+\frac{13}{7}>0\)

Và \(x^4+x^2-3x+3=x^4+\left(x-\frac{3}{2}\right)^2+\frac{3}{4}>0\)

\(\Rightarrow\left[{}\begin{matrix}x=-1\\x=0\end{matrix}\right.\)