\(\left\{{}\begin{matrix}x^2+y^2=1\\x^9+y^9=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\left(x+y\right)^2=1+2xy\\\left(x^3+y^3\right)^2=1+2x^3y^3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\left(x+y\right)^2=1+2xy\\\left(x+y\right)^2\left(x^2-xy+y^2\right)^2=1+2x^3y^3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\left(x+y\right)^2=1+2xy\\\left(x+y\right)^2\left[\left(x+y\right)^2-3xy\right]=1+2x^3y^3\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}\left(x+y\right)^2=1+2xy\\\left(1+2xy\right)\left(1+2xy-3xy\right)=1+2x^3y^3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\left(x+y\right)^2=1+2xy\\\left(1+2xy\right)\left(1-xy\right)=1+2x^3y^3\left(\text{1}\right)\end{matrix}\right.\)Giải (1) ta được \(\left[{}\begin{matrix}xy=0\Rightarrow x+y=\pm1\left(1'\right)\\xy=\frac{-1+\sqrt{3}}{2}\left(l\right)\\xy=\frac{-1-\sqrt{3}}{2}\Rightarrow x+y=\pm\sqrt[4]{3}\left(2'\right)\end{matrix}\right.\)
(1')=>\(\left(x;y\right)\in\left\{\left(0;1\right);\left(1;0\right);\left(0;-1\right);\left(-1;0\right)\right\}\)
(2')=> Hệ vô nghiệm
Thử lại: \(\left(x;y\right)\in\left\{\left(0;1\right);\left(1;0\right)\right\}\)thỏa mãn
Bài này có thể biện luận:
\(x^2+y^2=1\Rightarrow\left\{{}\begin{matrix}\left|x\right|\le1\\\left|y\right|\le1\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x^9\le x^2\\y^9\le y^2\end{matrix}\right.\)
\(\Rightarrow x^9+y^9\le x^2+y^2=1\)
Dấu "=" xảy ra khi và chỉ khi \(\left(x;y\right)=\left(1;0\right);\left(0;1\right)\)
