Question

\(\left(x-1\right)^{2018}+\left(x+3\right)^{2020}+\left|2x-y-z\right|=0\)\(0\)

Answer 1

Sửa đề 

\(\left(x-1\right)^{2018}+\left(y+3\right)^{2020}+\left|2x-y-z\right|=0\)

Vì \(\hept{\begin{cases}\left(x-1\right)^{2018}\ge0\forall x\\\left(y+3\right)^{2020}\ge0\forall y\\\left|2x-y-z\right|\ge0\forall x,y,z\end{cases}\Rightarrow\left(x-1\right)^{2018}+\left(y+3\right)^{2020}+\left|2x-y-z\right|\ge0\forall x,y,z}\)

Dấu " = " xảy ra khi :

( x - 1 )²⁰¹⁸ = 0 

=> x = 1 

( y + 3 )²⁰²⁰  = 0 

=> y = - 3 

Thay x = 1 ; y = -3 và | 2x - y - z | ta đc

| 2.1 + 3 - z | = 0 

=> | 5 - z | = 0

=> z = 5 

Vậy x = 1 ; y = -3 ; z = 5