a, \(\left(x+1\right)^8=16\left(x+1\right)^4\)
\(\Rightarrow\left(x+1\right)^8-16\left(x+1\right)^4=0\)
\(\Rightarrow\left(x+1\right)^4\left[\left(x+1\right)^4-16\right]=0\)
\(\Rightarrow\left[{}\begin{matrix}\left(x+1\right)^4=0\\\left(x+1\right)^4-16=0\end{matrix}\right.\)
+) \(\left(x+1\right)^4=0\Rightarrow x=-1\)
+) \(\left(x+1\right)^4-16=0\Rightarrow\left[{}\begin{matrix}x+1=2\\x+1=-2\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=1\\x=-3\end{matrix}\right.\)
Vậy x = -1 hoặc x = 1 hoặc x = -3
b, Ta có: \(\left\{{}\begin{matrix}\left(x-1\right)^2\ge0\\\left(y+1\right)^8\ge0\end{matrix}\right.\Rightarrow\left(x-1\right)^2+\left(y+1\right)^8\ge0\)
Mà \(\left(x-1\right)^2+\left(y+1\right)^8=0\)
\(\Rightarrow\left\{{}\begin{matrix}\left(x-1\right)^2=0\\\left(y+1\right)^8=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=1\\y=-1\end{matrix}\right.\)
Vậy x = 1 và y = -1
c, Ta có: \(\left\{{}\begin{matrix}\left(x-3\right)^2\ge0\\\left(y+1\right)^2\ge0\end{matrix}\right.\Rightarrow\left(x-3\right)^2+\left(y+1\right)^2\ge0\)
\(\Rightarrow\left(x-3\right)^2+\left(y+1\right)^2+1\ge1\)
Dấu " = " khi \(\left\{{}\begin{matrix}\left(x-3\right)^2=0\\\left(y+1\right)^2=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=3\\y=-1\end{matrix}\right.\)
Vậy \(MIN_{\left(x-3\right)^2+\left(y+1\right)^2+1}=1\) khi x = 3, y = -1
