\(a,A=x^2+2x+5=\left(x+1\right)^2+4\ge4\)
\(dấu"="\) \(xảy\) \(ra\Leftrightarrow x=-1\)
\(b,B=2x\left(x-3\right)=2x^2-6x=2\left(x^2-3x\right)=2\left(x^2-3x+\dfrac{9}{4}-\dfrac{9}{4}\right)=2\left[\left(x-\dfrac{3}{2}\right)^2-\dfrac{9}{4}\right]\ge-4,5\)
\(dấu"="\) \(xảy\) \(ra\Leftrightarrow x=\dfrac{3}{2}\)
\(c,C=x^2+y^2-x+6y+10\)
\(C=x^2-x+\dfrac{1}{4}+y^2+6y+9+0,75=\left(x-\dfrac{1}{2}\right)^2+\left(y+3\right)^2+0,75\ge0,75\)
\(dấu"="xảy\) \(ra\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{1}{2}\\y=-3\end{matrix}\right.\)
\(D=\left(x-1\right)\left(x+6\right)\left(x+2\right)\left(x+3\right)=\left(x^2+5x-6\right)\left(x^2+5x+6\right)\)
\(đặt\) \(x^2+5x-6=t\Rightarrow D=t\left(t+12\right)=t^2+12t=t^2+12t+36-36=\left(t-6\right)^2-36\ge-36\)
\(\Rightarrow MinD=-36\)
