Question

[Phạm Tường Vi]

/3x-5/-(2y+8)^20+(4z-3)^2018 bé hơn hoặc bằng 0

Answer 1

[Nguyệt]

\(\left|3x-5\right|-\left(2y+8\right)^{20}+\left(4z-3\right)^{2018}\le0\)

ta có:

\(\hept{\begin{cases}\left|3x-5\right|\ge0\\\left(2y+8\right)^{20}\ge0\\\left(4z-3\right)^{2018}\ge0\end{cases}}\Rightarrow\left|3x-5\right|-\left(2y+8\right)^{20}+\left(4z-3\right)^{2018}\ge0\)

mà \(\left|3x-5\right|-\left(2y+8\right)^{20}+\left(4z-3\right)^{2018}\le0\)=> \(\left|3x-5\right|-\left(2y+8\right)^{20}+\left(4z-3\right)^{2018}=0\)

=> \(\hept{\begin{cases}\left|3x-5\right|=0\\\left(2y+8\right)^{20}=0\\\left(4z-3\right)^{2018}=0\end{cases}}\Rightarrow\hept{\begin{cases}x=\frac{5}{3}\\y=-4\\z=\frac{4}{3}\end{cases}}\)

vậy \(x=\frac{5}{3},y=-4,z=\frac{4}{3}\)

Answer 2

[Phạm Tường Vi]

bạn nên có một bước giải thích vì sao 

(2y+8)\(\ge0\)

\(|3x-5|\ge0\)

\((4z-3)\ge0\)