Question

Giải hệ pt:

\(\left\{{}\begin{matrix}3x=\dfrac{x^2+2}{y^2}\\3y=\dfrac{y^2+2}{x^2}\end{matrix}\right.\)

Answer 1

ĐKXĐ: ...

\(\left\{{}\begin{matrix}3xy^2=x^2+2\\3x^2y=y^2+2\end{matrix}\right.\)

Chia vế cho vế:

\(\dfrac{y}{x}=\dfrac{x^2+2}{y^2+2}\Rightarrow y^3+2y=x^3+2x\)

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

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

\(\Leftrightarrow x=y\)

Thế vào pt đầu:

\(3x^3=x^2+2\Leftrightarrow3x^3-x^2-2=0\)

\(\Leftrightarrow\left(x-1\right)\left(3x^2+2x+2\right)=0\)

\(\Leftrightarrow x=1\Rightarrow y=1\)