Question

Giải hệ :

\(\left\{{}\begin{matrix}x^2\left(4y+1\right)-2y=-3\\x^2\left(x^2-12y\right)+4y^2=9\end{matrix}\right.\)

Answer 1

hpt \(\Leftrightarrow\left\{{}\begin{matrix}x^2\left(4y+1\right)=2y-3\\x^2\left(x^2-12y\right)=-4y^2+9\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x^2\left(4y+1\right)\left(2y+3\right)=4y^2-9\\x^2\left(x^2-12y\right)=-4y^2+9\end{matrix}\right.\)

Cộng theo vế 2 pt ta đc:

\(x^2\left(x^2+8y^2+2y+3\right)=0\)

\(\Leftrightarrow x^2\left[x^2+7y^2+\left(y+1\right)^2+2\right]=0\)

\(\Leftrightarrow x=0\)

\(\Rightarrow y=\dfrac{3}{2}\left(tm\right)\)