Ta có:
\(A=2x^2-12x+20=2\left(x^2-6x+10\right)=2\left(x-3\right)^2+2\)
Vì \(2\left(x-2\right)^2\ge0\forall x\in R\Rightarrow2\left(x-3\right)^2+2\ge2\forall x\in R\)
\(\Rightarrow A_{min}=2\Leftrightarrow x-3=0\Leftrightarrow x=3\)
Vậy, A đạt GTNN là 2 <=> x = 3
\(2x^2-12x+20\\ =2\left(x^2-6x+10\right)\\ =2\left(x^2-6x+9+1\right)\\ =2\left(x^2-6x+9\right)+2\\ =2\left(x-3\right)^2+2\\ \)
\(2\left(x-3\right)^2\ge0\\ \Rightarrow2\left(x-3\right)^2+2\ge2\\ \Rightarrow2x^2-12x+20\ge2\\\)
Nếu 2x2-12x+20 \(\ge\)2=>2(x-3)2=0
=>(x-3)2=0
=>x-3=0
=>x=3
Vậy 2x2 -12x +20 có GTNN là 2 nếu x=3
