ĐKXĐ: ...
\(\Leftrightarrow\left\{{}\begin{matrix}2x^2y=y^2+1\\2xy^2=x^2+1\end{matrix}\right.\)
Chia vế cho vế: \(\frac{x}{y}=\frac{y^2+1}{x^2+1}\Leftrightarrow x^3+x=y^3+y\)
\(\Rightarrow x^3-y^3+x-y=0\Leftrightarrow\left(x-y\right)\left(x^2+xy+y^2+1\right)=0\)
\(\Leftrightarrow x=y\)
\(\Rightarrow2x^3=x^2+1\Leftrightarrow2x^3-x^2-1=0\)
\(\Leftrightarrow\left(x-1\right)\left(2x^2+x+1\right)=0\Rightarrow x=1\Rightarrow y=1\)
ĐKXĐ:...
\(\Leftrightarrow\left\{{}\begin{matrix}2x^2-y=\frac{1}{y}\\2y^2-x=\frac{1}{x}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x^2y-y^2=1\\2xy^2-x^2=1\end{matrix}\right.\Rightarrow2xy\left(x-y\right)+\left(x^2-y^2\right)=0\)
\(\Leftrightarrow\left(x-y\right)\left(x+y\right)+2xy\left(x-y\right)=0\)
\(\Leftrightarrow\left(x-y\right)\left(x+y+2xy\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=y\\x+y+2xy=0\end{matrix}\right.\)
Nhẩm \(pt:x+y+2xy\) có nghiệm x=y=-1
Rồi đến đây cậu tự lm nốt nhé
