Question

cho hệ phương trình:

x2+y2=5

x4-x2y2+y4=13

Answer 1

\(\left\{{}\begin{matrix}x^2+y^2=5\\x^4-x^2y^2+y^4=13\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\left(x+y\right)^2-2xy=5\\\left(x^2+y^2\right)^2+x^2y^2=13\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}\left(x+y\right)^2-2xy=5\\\left[\left(x+y\right)^2-2xy\right]^2+x^2y^2=13\end{matrix}\right.\)

Đặt S=x+y; P=xy( \(S^2\ge4P\))

\(\Rightarrow\left\{{}\begin{matrix}S^2-2P=5\\\left[S^2-2P\right]^2+P^2=13\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}S^2-2P=5\\P^2=13-25=-12\left(vl\right)\end{matrix}\right.\)

Vậy hpt vô nghiệm.