Question

Giải hệ phương trình :\(\left\{{}\begin{matrix}x^2+4y^2-5=0\\4x^2y+8xy^2+5x+10y-1=0\end{matrix}\right.\)

Answer 1

\(\left\{{}\begin{matrix}\left(x+2y\right)^2-4xy=5\\4xy\left(x+2y\right)+5\left(x+2y\right)=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a^2-4b=5\\4ab+5a=1\end{matrix}\right.\)

\(\left\{{}\begin{matrix}4b=a^2-5\\a\left(a^2-5\right)+5a=1\end{matrix}\right.\)

\(\Rightarrow a^3=1\)=> a=1 => 4b= 1 -5 =4=> b = -1

=>\(\left\{{}\begin{matrix}x+2y=1\\xy=-1\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}y=1\\x=-1\end{matrix}\right.\\\left\{{}\begin{matrix}y=-\dfrac{1}{2}\\x=2\end{matrix}\right.\end{matrix}\right.\)