Bài 4:
\(A=\left(x^2-9\right)^2+\left|y-2\right|+10\)
\(\text{Ta có: }\left(x^2-9\right)^2\ge0\text{ với }\forall x\in R\left(1\right)\)
\(\left|y-2\right|\ge0\text{ với }\forall y\in R\left(2\right)\)
\(\text{Từ (1) và (2) }\Rightarrow\left(x^2-9\right)^2+\left|y-2\right|\ge0\)
\(\Rightarrow\left(x^2-9\right)^2+\left|y-2\right|+10\ge10\)
\(\text{Dấu "=" xảy ra }\Leftrightarrow\left(x^2-9\right)^2+\left|y-2\right|+10=10\)
\(\Rightarrow\left\{{}\begin{matrix}\left(x^2-9\right)^2=0\\\left|y-2\right|=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x^2-9=0\\y-2=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x^2=9\\y=2\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}\Rightarrow\left[{}\begin{matrix}x=3\\x=-3\end{matrix}\right.\\y=2\end{matrix}\right.\)
\(\Rightarrow\left(x,y\right)\in\left\{\left(-3;2\right);\left(3;2\right)\right\}\)
\(\text{Vậy }Min_A=10\text{ tại }\left(x,y\right)\in\left\{\left(-3;2\right);\left(3;2\right)\right\}\)
