mình nghĩ đề sai, chắc đề vậy mới đúng :))
\(y=\dfrac{2x+1}{x^2+2x+3}=\dfrac{-x^2-2x-3+x^2+4x+4}{x^2+2x+3}\)
\(y=\dfrac{x^2+4x+4}{x^2+2x+3}-1=\dfrac{\left(x+2\right)^2}{\left(x+1\right)^2+1}-1\ge-1\forall x\in R\)
dấu '=' xảy ra khi \(x+2=0\Leftrightarrow x=-2\)
vậy \(y_{MIN}=-1\) khi x=-2
\(y=\dfrac{2x+1}{x^2+2x+3}=\dfrac{4x+2}{2\left(x^2+2x+3\right)}\)
\(y=\dfrac{x^2+2x+3-x^2+2x-1}{2\left(x^2+2x+3\right)}\)
\(y=\dfrac{-x^2+2x-1}{2\left(x^2+2x+3\right)}+\dfrac{1}{2}\)
\(y=\dfrac{-\left(x-1\right)^2}{2\left(x+1\right)^2+2}+\dfrac{1}{2}\le\dfrac{1}{2}\forall x\in R\)
dấu '=' xảy ra khi \(x-1=0\Leftrightarrow x=1\)
vậy \(y_{max}=\dfrac{1}{2}\) khi x=1
\(Y=\dfrac{2x-1}{x^2+2x+3}\Leftrightarrow x^2.Y+x.\left(2Y-2\right)+3Y+1=0\)
\(\Delta'=\left(Y-1\right)^2-Y\left(3Y+1\right)\ge0\)
\(\Leftrightarrow\dfrac{-3-\sqrt{17}}{4}\le Y\le\dfrac{-3+\sqrt{17}}{4}\)
