Question

Answer 1

\(\left\{{}\begin{matrix}\left|3x-5\right|>=0\\\left(2y+5\right)^{208}>=0\\\left(4z-3\right)^{20}>=0\end{matrix}\right.\)

Do đó: \(\left|3x-5\right|+\left(2y+5\right)^{208}+\left(4z-3\right)^{20}>=0\forall x,y,z\)

Dấu '=' xảy ra khi x=5/3; y=-5/2; z=3/4