\(HPT\Leftrightarrow\left\{{}\begin{matrix}3x^2+3y^3=15\\2x^2-3y^3=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}5x^2=20\\x^2+y^3=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x^2=4\\y^3=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x=2\\x=-2\end{matrix}\right.\\y=1\end{matrix}\right.\)
Vậy \(\left(x;y\right)=\left(2;1\right);\left(-2;1\right)\)
Đúng 2
Bình luận (0)
