Sửa đề:
Tìm x;y;z biết\(\left|3x-5\right|+\left(2y-8\right)^{20}+\left(4z-3\right)^{2018}\le0\)
Ta có: \(\hept{\begin{cases}\left|3x-5\right|\ge0\forall x\\\left(2y-8\right)^{20}\ge0\forall y\\\left(4z-3\right)^{2018}\ge0\forall z\end{cases}}\)
\(\Rightarrow\left|3x-5\right|+\left(2y-8\right)^{20}+\left(4z-3\right)^{2018}\ge0\)
Mà \(\left|3x-5\right|+\left(2y-8\right)^{20}+\left(4z-3\right)^{2018}\le0\)
\(\Rightarrow\left|3x-5\right|+\left(2y-8\right)^{20}+\left(4z-3\right)^{2018}=0\)
\(\Rightarrow\hept{\begin{cases}\left|3x-5\right|=0\\\left(2y-8\right)^{20}=0\\\left(4z-3\right)^{2018}=0\end{cases}}\Leftrightarrow\hept{\begin{cases}3x-5=0\\2y-8=0\\4z-3=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=\frac{5}{3}\\y=4\\z=\frac{3}{4}\end{cases}}\)
Vậy \(\hept{\begin{cases}x=\frac{5}{3}\\y=4\\z=\frac{3}{4}\end{cases}}\)
Tham khảo nhé~
