Question

Giải hệ phương trình:

\(\left\{{}\begin{matrix}x^2+x=5+y^2+y\\x^3+y^3=x^2y+xy^2+6\end{matrix}\right.\)

Answer 1

\(hpt\Leftrightarrow\left\{{}\begin{matrix}x^2-y^2=5-\left(x-y\right)\\\left(x+y\right)\left(x^2+y^2-xy\right)=xy\left(x-y\right)+6\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x^2-y^2=5-\left(x-y\right)\\\left(x^2-y^2\right)\left(x-y\right)=6\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x^2-y^2=5-\left(x-y\right)\\x^2-y^2=\frac{6}{x-y}\end{matrix}\right.\)

\(\Rightarrow5-\left(x-y\right)=\frac{6}{x-y}\)\(\Rightarrow x-y=3\)

\(\Rightarrow x+y=\frac{6}{\left(x-y\right)^2}=\frac{6}{9}=\frac{2}{3}\)

\(\Rightarrow\left\{{}\begin{matrix}x=\frac{11}{6}\\y=-\frac{7}{6}\end{matrix}\right.\)