Question

Giải hệ:

\(\left\{{}\begin{matrix}x^4+y^6+z^8\le1\\x^{2017}+y^{2019}+z^{2021}\ge1\end{matrix}\right.\)

Answer 1

Ta có \(-1\le x,yz\le1\)
\(=>x^{2007}+y^{2009}+z^{2011}\ge x^6+y^8+z^{10}\)
\(< =>x^6\left(1-x^{2001}\right)+y^8\left(1-y^{2001}\right)+z^{10}\left(1-z^{2001}\right)\le0\)
Từ \(-1\le x,y,z\le1\) ta thấy
\(x^6\left(1-x^{2001}\right),y^8\left(1-y^{2001}\right),z^{10}\left(1-y^{2001}\right)\ge0\)
Do đó \(< =>x^6\left(1-x^{2001}\right)=y^8\left(1-y^{2001}\right)=z^{10}\left(1-z^{2001}\right)=0\)
\(< =>x,y,z=1\left(x,y,z=0\right)\)
\(=>\left(x;y;z\right)=\left(1;0;0\right),\left(0;1;0\right),\left(0;0;1\right)\)