Question

giải hệ phương trình:
\(\left\{{}\begin{matrix}x^3+4y=y^3+16x\\1+y^2=5\left(1+x^2\right)\end{matrix}\right.\)

Answer 1

\(\Leftrightarrow\left\{{}\begin{matrix}x^3-y^3=16x-4y\\-4=5x^2-y^2\end{matrix}\right.\)

\(\Rightarrow-4\left(x^3-y^3\right)=4\left(4x-y\right)\left(5x^2-y^2\right)\)

\(\Leftrightarrow y^3-x^3=20x^3-4xy^2-5x^2y+y^3\)

\(\Leftrightarrow21x^3-5x^2y-4xy^2=0\)

\(\Leftrightarrow x\left(21x^2-5xy-4y^2\right)=0\)

\(\Leftrightarrow x\left(7x-4y\right)\left(3x+y\right)=0\)

\(\Rightarrow...\)