b) Ta có :
\(\left(x-8\right)^2\ge0;\forall x\)
\(\Rightarrow\left(x-8\right)^2+2019\ge2019;\forall x\)
Hay\(B\ge2019;\forall x\)
Dấu "=" xảy ra \(\Leftrightarrow\left(x-8\right)^2=0\)
\(\Leftrightarrow x=8\)
Vậy MIN B=2019 \(\Leftrightarrow x=8\)
c) Vì \(\hept{\begin{cases}-|20-x|\le0;\forall x\\-|70+y|\le0;\forall y\end{cases}}\)
\(\Rightarrow-|20-x|-|70+y|\le0;\forall x,y\)
\(\Rightarrow90-|20-x|-|70+y|\le90-0;\forall x,y\)
Hay \(C\le90;\forall x,y\)
Dấu "=" xảy ra \(\Leftrightarrow\hept{\begin{cases}|20-x|=0\\|70+y|=0\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x=20\\y=-70\end{cases}}\)
Vậy MAX C=90 \(\Leftrightarrow\hept{\begin{cases}x=20\\y=-70\end{cases}}\)
A=|2-2|+|y+5|-10
\(=\left|y+5\right|-10\ge-10\forall x\in R\)
Dấu "=" xảy ra <=> \(\left|y+5\right|=0\Leftrightarrow y+5=0\Leftrightarrow y=-5\)
Vậy Amin =-10 tại y=-5