\(a,Taco:\)
\(\left(x-1\right)^2,\left(y-3\right)^8\ge0\)
\(\Rightarrow\left(x-1\right)^2+\left(y-3\right)^8=0\Leftrightarrow\hept{\begin{cases}x-1=0\Leftrightarrow x=1\\y-3=0\Leftrightarrow y=3\end{cases}}\)
\(b,Taco:\)
\(|x-2018|+\left(y-2019\right)^{2018}\ge0\)
\(\Rightarrow|x-2018|+\left(y-2019\right)^{2018}=0\Leftrightarrow\hept{\begin{cases}x-2018=0\Leftrightarrow x=2018\\y-2019=0\Leftrightarrow y=2019\end{cases}}\)
\(a,\left(x-1\right)^2+\left(y-3\right)^8=0\)
Vì \(\left(x-1\right)^2\ge0vs\forall x;\left(y-3\right)^8\ge0vs\forall y\)
\(\Rightarrow\hept{\begin{cases}\left(x-1\right)^2=0\\\left(y-3\right)^8=0\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}x-1=0\\y-3=0\end{cases}}\) \(\Rightarrow\hept{\begin{cases}x=1\\y=3\end{cases}}\)
Vậy x = 1, y = 3
\(b,\left|x-2018\right|+\left(y-2019\right)^{2018}=0\)
Vì \(\left|x-2018\right|\ge0vs\forall x;\left(y-2019\right)^{2018}\ge0vs\forall y\)
\(\Rightarrow\hept{\begin{cases}\left|x-2018\right|=0\\\left(x-2019\right)^{2018}=0\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}x-2018=0\\x-2019=0\end{cases}}\) \(\Rightarrow\hept{\begin{cases}x=2018\\y=2019\end{cases}}\)
Vậy x = 2018; y = 2019
Ta có: \(\hept{\begin{cases}\left(x-1\right)^2\ge0\forall x\\\left(y-3\right)^8\ge0\forall y\end{cases}}\)
\(\Rightarrow\left(x-1\right)^2+\left(y-3\right)^8\ge0\forall x;y\)
Mà \(\left(x-1\right)^2+\left(y-3\right)^8=0\)
\(\Rightarrow\hept{\begin{cases}\left(x-1\right)^2=0\\\left(y-3\right)^8=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x-1=0\\y-3=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=1\\y=3\end{cases}}}\)
Vậy \(\hept{\begin{cases}x=1\\y=3\end{cases}}\)
