Question

tìm x,y,z biết: \(\left(3x+1\right)^{2018}+\left(2y-1\right)^{2018}+\left|x+2y-z\right|^{2018}=0\)

Answer 1

Vì : (3x+1)²⁰¹⁸+(2y-1)²⁰¹⁸+\(\left|x+2y-z\right|\)²⁰¹⁸=0

Nên: \(\left\{{}\begin{matrix}\left(3x+1\right)^{2018}=0\\\left(2y-1\right)^{2018}\\\left|x+2y-z\right|^{2018}=0\end{matrix}\right.=0\) ⇔\(\left\{{}\begin{matrix}3x+1=0\\2y-1=0\\x+2y-z=0\end{matrix}\right.\)

⇔\(\left\{{}\begin{matrix}x=\dfrac{-1}{3}\\y=\dfrac{1}{2}\\\dfrac{-1}{3}+1-z=0\end{matrix}\right.\) ⇔\(\left\{{}\begin{matrix}x=\dfrac{-1}{3}\\y=\dfrac{1}{2}\\z=\dfrac{2}{3}\end{matrix}\right.\)

Vậy : x=\(\dfrac{-1}{3}\) ; y=\(\dfrac{1}{2}\) ; z=\(\dfrac{2}{3}\)