c)\(C=2x^2+y^2+2xy-8x+2019\)
\(=\left(x^2+2xy+y^2\right)+\left(x^2-8x+16\right)+2003\)
\(=\left(x+y\right)^2+\left(x-4\right)^2+2003\)
Vì \(\left(x+y\right)^2\ge0,\forall x,y\)
\(\left(x-4\right)^2\ge0,\forall x\)
Nên \(\left(x+y\right)^2+\left(x-4\right)^2\ge0\)
\(\Rightarrow\left(x+y\right)^2+\left(x-4\right)^2+2003\ge2003\)
\(\Rightarrow min_C=2003\) khi \(\left\{{}\begin{matrix}x-4=0\\x+y=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=4\\y=-4\end{matrix}\right.\)
d)\(D=\left(x-1\right)\left(x+2\right)\left(x+3\right)\left(x+6\right)\)
\(=\left[\left(x-1\right)\left(x+6\right)\right]\left[\left(x+2\right)\left(x+3\right)\right]\)
\(=\left(x^2+5x-6\right)\left(x^2+5x+6\right)\)
Đặt \(t=x^2+5x\) ta được:
\(D=\left(t-6\right)\left(t+6\right)=t^2-36\)
Vì \(t^2\ge0,\forall t\)
\(\Rightarrow t^2-36\ge-36\)
\(\Rightarrow min_D=-36\) khi \(t=0\)
\(\Leftrightarrow x^2-5x=0\)
\(\Leftrightarrow x\left(x-5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=5\end{matrix}\right.\)
Vậy Min D = -36 khi \(x\in\left\{0;5\right\}\)
