( x - 1 )2018 + ( y + 3 )2020 + ( z - 5 )2022 = 0
Ta thấy : ( x - 1 )2018 \(\ge0\) ; ( y + 3 )2020 \(\ge0\) ; ( z - 5 )2022 \(\ge0\)
\(\Rightarrow\left(x-1\right)^{2018}+\left(y+3\right)^{2020}+\left(z-5\right)^{2022}\ge0\)
Theo đề,ta có : \(\left(x-1\right)^{2018}=\left(y+3\right)^{2020}=\left(z-5\right)^{2022}=0\)
+) \(\left(x-1\right)^{2018}=0\Rightarrow x-1=0\Rightarrow x=1\)
+) \(\left(y+3\right)^{2020}=0\Rightarrow y+3=0\Rightarrow y=-3\)
=) \(\left(z-5\right)^{2022}=0\Rightarrow z-5=0\Rightarrow z=5\)
Vậy : x = 1 ; y = -3 ; z = 5
\(\text{Ta có:}\)
\(\hept{\begin{cases}\left(x-1\right)^{2018}\ge0\\\left(y+3\right)^{2020}\ge0\\\left(z-5\right)^{2022}\ge0\end{cases}}\text{mà:}\left(x-1\right)^{2018}+\left(y-2\right)^{2020}+\left(z-3\right)^{2022}=0\text{ nên:}\)
\(\hept{\begin{cases}\left(x-1\right)^{2018}=0\\\left(y+3\right)^{2018}=0\\\left(z-5\right)^{2018}=0\end{cases}}\Rightarrow\hept{\begin{cases}x=1\\y=-3\\z=5\end{cases}}\)
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