\(\left(x-3\right)^{24}>=0\forall x\)
\(\left|y-5\right|^{23}>=0\forall y\)
\(\sqrt{\left(x+y-z\right)}^{22}>=0\forall x,y,z\)
Do đó: \(\left(x-3\right)^{24}+\left|y-5\right|^{23}+\sqrt{\left(x+y-z\right)^{22}}>=0\forall x,y,z\)
Dấu '=' xảy ra khi \(\left\{{}\begin{matrix}x-3=0\\y-5=0\\x+y-z=0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=3\\y=5\\z=x+y=8\end{matrix}\right.\)
\(M=2x^4+3y^3-z^2+xyz\)
\(=2\cdot3^4+3\cdot5^3-8^2+3\cdot5\cdot8\)
=162+375-64+120
=593