Question

Giải hệ pt

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

Answer 1

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

Nhân vế với vế:

\(-4\left(x^3-y^3\right)=\left(16x-4y\right)\left(5x^2-y^2\right)\)

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

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

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

\(\Rightarrow\left[{}\begin{matrix}x=0\\y=\dfrac{7x}{4}\\y=-3x\end{matrix}\right.\) thế xuống pt dưới:

\(\Rightarrow\left[{}\begin{matrix}1+y^2=5\\1+\left(\dfrac{7x}{4}\right)^2=5\left(1+x^2\right)\\1+9x^2=5\left(1+x^2\right)\end{matrix}\right.\) \(\Leftrightarrow...\)