Question

Giải hệ phương trình 3/x²=2x+y,3/y²=2y+x

Answer 1

ĐKXĐ: ...

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

\(\Rightarrow2\left(x^3-y^3\right)+xy\left(x-y\right)=0\)

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x-y=0\\2x^2+2y^2+3xy=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=y\\2\left(x+\frac{3}{4}y\right)^2+\frac{7y^2}{8}=0\end{matrix}\right.\)

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

Với \(x=y\) thay vào pt đầu: \(3x^3=3\Rightarrow x=1\Rightarrow y=1\)