Question

\(\left\{{}\begin{matrix}\frac{x^2}{y}+x=2\\\frac{y^2}{x}+y=\frac{1}{2}\end{matrix}\right.\)

Answer 1

\(\left\{{}\begin{matrix}\frac{x^2}{y}+x=2\left(1\right)\\\frac{y^2}{x}+y=\frac{1}{2}\left(2\right)\end{matrix}\right.\)

\(\left(1\right)-\left(2\right)=\frac{x^2}{y}-\frac{y^2}{x}+\left(x-y\right)=\frac{3}{2}\)

\(\Leftrightarrow\frac{x^2-y^2}{y}-\frac{y^2-x^2}{x}-\frac{3}{2}=0\)

\(\Leftrightarrow2x\left(x^2-y^2\right)+2y\left(x^2-y^2\right)-2xy=0\)

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

Ta thấy x=y=0 ko là nghiệm.

pt=>\(\frac{x^2-y^2}{y}+\frac{x^2-y^2}{x}=1\)

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

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

\(\Rightarrow\left[{}\begin{matrix}x=y\\x=-y\end{matrix}\right.\)

Mong đừng sai TvT

#Walker