Ta có : \(\left\{{}\begin{matrix}x^2-y^2+x-y=5\\x^3-x^2y-xy^2+y^3=6\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}\left(x-y\right)\left(x+y\right)+\left(x-y\right)=5\\x^2\left(x-y\right)-y^2\left(x-y\right)=6\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}\left(x-y\right)\left(x+y+1\right)=5\\\left(x-y\right)\left(x-y\right)\left(x+y\right)=6\end{matrix}\right.\)
- Đặt \(x-y=a,x+y=b\) ta được hệ phương trình :
\(\left\{{}\begin{matrix}a\left(b+1\right)=5\\a^2b=6\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}ab+a=5\\ab.a=6\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}ab=5-a\\a\left(5-a\right)=6\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}ab=5-a\\-a^2+5a-6=0\left(I\right)\end{matrix}\right.\)
- Giair phương trình ( I ) ta được : \(\left[{}\begin{matrix}a=3\\a=2\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}b=\frac{2}{3}\\b=\frac{3}{2}\end{matrix}\right.\)
- Thay lại \(a=3,b=\frac{2}{3}\) vào lại ta được : \(\left\{{}\begin{matrix}x-y=3\\x+y=\frac{2}{3}\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}x=\frac{11}{6}\\y=-\frac{7}{6}\end{matrix}\right.\)
- Thay lại \(a=2,b=\frac{3}{2}\) vào lại ta được : \(\left\{{}\begin{matrix}x-y=2\\x+y=\frac{3}{2}\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}x=\frac{7}{4}\\y=-\frac{1}{4}\end{matrix}\right.\)
Vây ....
