\(E=\dfrac{x^2+x+1}{x^2+1}\)
\(\Leftrightarrow E\left(x^2+1\right)=x^2+x+1\)
\(\Leftrightarrow Ex^2+E-x^2-x-1=0\)
\(\Leftrightarrow\left(E-1\right)x^2-x+E-1=0\)
\(\left(a=E-1,b=-1,c=E-1\right)\)
* \(E\ge0\)
\(\Rightarrow\Delta=b^2-4ac\ge0\)
\(\Rightarrow\left(-1\right)^2-4.\left(E-1\right)^2\ge0\)
\(\Leftrightarrow1^2-4E^2+8E-4\ge0\)
\(\Leftrightarrow-4E^2+8E-3\ge0\)
\(\Leftrightarrow\dfrac{1}{2}\le E\le\dfrac{3}{2}\)
\(\Rightarrow Min_E=\dfrac{1}{2}\Leftrightarrow x=\dfrac{-b}{2a}=\dfrac{1}{2.\left(\dfrac{1}{2}-1\right)}=-1\)
\(Max_E=\dfrac{3}{2}\Leftrightarrow x=\dfrac{-b}{2a}=\dfrac{1}{2.\left(\dfrac{3}{2}-1\right)}=1\)
\(=\dfrac{-\left(x^2-2x+1\right)+3\left(x^2+1\right)}{2\left(x^2+1\right)}=\dfrac{-\left(x-1\right)^2}{2\left(x^2+1\right)}+\dfrac{3}{2}\le\dfrac{3}{2}\forall x\)
Dấu "=" xảy ra khi"
\(x-1=0\\ \Leftrightarrow x=1\)
\(\text{* }E=\dfrac{x^2+x+1}{x^2+1}=\dfrac{2x^2+2x+2}{2\left(x^2+1\right)}\\ =\dfrac{\left(x^2+2x+1\right)+\left(x^2+1\right)}{2\left(x^2+1\right)}=\dfrac{\left(x+1\right)^2}{2\left(x^2+1\right)}+\dfrac{1}{2}\ge\dfrac{1}{2}\forall x\)
Dấu \("="\) xảy ra khi:
\(x-1=0\\ \Leftrightarrow x=1\)
\(\text{* }E=\dfrac{x^2+x+1}{x^2+1}=\dfrac{2x^2+2x+2}{2\left(x^2+1\right)}\\ =\dfrac{-\left(x^2-2x+1\right)+2\left(x^2+1\right)}{2\left(x^2+1\right)}=\dfrac{-\left(x-1\right)^2}{2\left(x^2+1\right)}+2\le2\forall x\)
Dấu \("="\) xảy ra khi:
\(x-1=0\\ \Leftrightarrow x=1\)
Vậy..........
