\(y=\dfrac{x^2+x+1}{x^2+2x+2}=\dfrac{\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}}{\left(x+1\right)^2}\ge\dfrac{3}{4}\)
Dấu "=" xảy ra khi: \(x=-\dfrac{1}{2}\)
Giải xàm vc
\(y=\dfrac{x^2+x+1}{x^2+2x+2}\)
\(\Leftrightarrow x^2\left(y-1\right)+x\left(2y-1\right)+\left(2y-1\right)=0\)
\(\Delta=\left(2y-1\right)^2-4\left(y-1\right)\left(2y-1\right)\)
\(=-\left(2y-1\right)\left(2y-3\right)\)
Pt có nghiêm khi \(\left(2y-1\right)\left(2y-3\right)\le0\)
\(\Leftrightarrow\left\{{}\begin{matrix}2y-1\ge0\\2y-3\le0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}y\ge\dfrac{1}{2}\\y\le\dfrac{3}{2}\end{matrix}\right.\)
"=" <=> \(x=0\)
