A = 2x2 +4x
⇒ A= 2x2 +4x + 2 - 2
⇒A = 2(x2 +2x + 1) - 2
⇒A = 2(x+1)2 -2 ≥ -2
Dấu "=" xảy ra khi và chỉ khi:
(x+1)2 = 0
⇒ x + 1 = 0 ⇒ x = -1
\(2x^2+4x\)
\(=2\left(x^2+x\right)=2\left(x^2+2x\dfrac{1}{2}+\dfrac{1}{4}\right)-\dfrac{1}{2}\)
\(=2\left(x+\dfrac{1}{2}\right)^2-\dfrac{1}{2}\)
\(2\left(x+\dfrac{1}{2}\right)^2\ge0\)\(\Rightarrow2\left(x+\dfrac{1}{2}\right)^2-\dfrac{1}{2}\ge\dfrac{-1}{2}\)
\(\Rightarrow min=\dfrac{-1}{2}\Leftrightarrow x-\dfrac{1}{2}=0\Rightarrow x=\dfrac{1}{2}\)
mk sai lm lại nek
\(2x^2+4x=2\left(x^2+2x+1\right)-2=2\left(x+1\right)^2-1\)
\(2\left(x^2+1\right)\ge0\Rightarrow2\left(x^2+1\right)-2\ge-2\)\(\Rightarrow min=-2\Leftrightarrow x+1=0\Rightarrow x=-1\)
