Question

(x-1)²⁰¹⁸+(2y-1)²⁰¹⁸+|x+2y-z|²⁰¹⁹=0

Answer 1

Ta có: \(\left(x-1\right)^{2018}>=0\forall x\)

\(\left(2y-1\right)^{2018}>=0\forall y\)

\(\left|x+2y-z\right|^{2019}>=0\forall x,y,z\)

Do đó: \(\left(x-1\right)^{2018}+\left(2y-1\right)^{2018}+\left|x+2y-z\right|^{2019}>=0\forall x,y,z\)

Dấu '=' xảy ra khi \(\left\{{}\begin{matrix}x-1=0\\2y-1=0\\x+2y-z=0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x=1\\y=\dfrac{1}{2}\\z=x+2y=1+2\cdot\dfrac{1}{2}=1+1=2\end{matrix}\right.\)