\(A=2x^2-4x-6\)
\(=\left(2x^2-4x+2\right)-8\)
\(=2\left(x^2-2x+1\right)-8\)
\(=2\left(x-1\right)^2-8\)
Ta có :
\(2\left(x-1\right)^2\ge0\forall x\)
\(\Rightarrow2\left(x-1\right)^2-8\ge-8\forall x\)
Dấu = xảy ra \(\Leftrightarrow2\left(x-1\right)^2=0\)
\(\Leftrightarrow\left(x-1\right)^2=0\)
\(\Leftrightarrow x-1=0\)
\(\Leftrightarrow x=1\)
Vậy \(Min_A=-8\Leftrightarrow x=1\)
\(A=2x^2-4x-6\\ =2x^2-4x+2-8\\ =\left(2x^2-4x+2\right)-8\\ =2\left(x^2-2x+1\right)-8\\ =2\left(x-1\right)^2-8\\ Do\text{ }\left(x-1\right)^2\ge0\forall x\\ \Rightarrow2\left(x-1\right)^2\ge0\forall x\\ A=2\left(x-1\right)^2-8\ge-8\forall x\)
Dấu "=" xảy ra khi:
\(\left(x-1\right)^2=0\\ \Leftrightarrow x-1=0\\ \Leftrightarrow x=1\)
Vậy \(A_{\left(Min\right)}=-8\) khi \(x=1\)
