Question

Giải hệ pt: \(\left\{{}\begin{matrix}x^3+8y^3-4xy^2=1\\2x^4+8y^4-2x-y=0\end{matrix}\right.\)

Answer 1

\(\Leftrightarrow\left\{{}\begin{matrix}x^3+8y^3-4xy^2=1\\2x+y=2x^4+8y^4\end{matrix}\right.\)

Nhân vế với vế:

\(\left(2x+y\right)\left(x^3+8y^3-4xy^2\right)=2x^4+8y^4\)

\(\Leftrightarrow12xy^3-8x^2y^2+x^3y=0\)

\(\Leftrightarrow xy\left(12y^2-2xy+x^2\right)=0\)

\(\Leftrightarrow xy=0\Rightarrow\left[{}\begin{matrix}x=0\Rightarrow y=...\\y=0\Rightarrow x=...\end{matrix}\right.\)