Question

giải hệ phương trình

\(X^3+Y^3=7\)

\(XY\left(X+Y\right)=-2\)

Answer 1

\(\left\{{}\begin{matrix}X^3+Y^3=7\left(1\right)\\XY\left(X+Y\right)=-2\left(2\right)\end{matrix}\right.\)

(1)\(\Leftrightarrow\left(X+Y\right)\left(X^2-XY+Y^2\right)=7\Leftrightarrow\left(X+Y\right)\left[\left(X+Y\right)^2-3XY\right]=7\Leftrightarrow\left(X+Y\right)^3-3XY\left(X+Y\right)=7\)

Đặt t=XY;z=X+Y

Vậy phương trình bây giờ là \(\left\{{}\begin{matrix}t^3-3tz=7\\tz=-2\end{matrix}\right.\)\(\Leftrightarrow\)\(\left\{{}\begin{matrix}t=1\\z=-2\end{matrix}\right.\)\(\Leftrightarrow\)\(\left\{{}\begin{matrix}XY=1\\X+Y=-2\end{matrix}\right.\)\(\Leftrightarrow\)\(\left[{}\begin{matrix}\left\{{}\begin{matrix}X=1+\sqrt{2}\\Y=1-\sqrt{2}\end{matrix}\right.\\\left\{{}\begin{matrix}X=1-\sqrt{2}\\Y=1+\sqrt{2}\end{matrix}\right.\end{matrix}\right.\)

Vậy (X;Y)={(\(1+\sqrt{2};1-\sqrt{2}\));(\(1-\sqrt{2};1+\sqrt{2}\))}