Ta có:
\(y=\dfrac{4x+3}{x^2+1}\)
\(\Rightarrow yx^2+y=4x+3\)
\(\Rightarrow yx^2-4x+y-3=0\)
* \(y=0\Rightarrow x=-\dfrac{3}{4}\)
* \(y\ge0\Rightarrow\Delta=b^2-4ac\ge0\)
\(\Rightarrow\left(-4\right)^2-4y.\left(y-3\right)\ge0\)
\(\Rightarrow16-4y^2+12y\ge0\)
⇒ 1≤ y ≤ 4
\(\Rightarrow Min_y=-1\Leftrightarrow x=-\dfrac{b}{2a}=\dfrac{4}{2.\left(-1\right)}=-2\)
\(Max_y=4\Leftrightarrow x=\dfrac{-b}{2a}=\dfrac{4}{2.4}=\dfrac{1}{2}\)
\(y=\dfrac{4x+3}{x^2+1}=\dfrac{-x^2+4x-4+x^2+1}{x^2+1}=\dfrac{-\left(x^2-4x+4\right)}{x^2+1}+\dfrac{x^2+1}{x^2+1}=\dfrac{-\left(x-2\right)^2}{x^2+1}+1\)vì x2\(\ge0\Leftrightarrow x^2+1\ge0\)
(x-2)2\(\ge0\Leftrightarrow-\left(x-2\right)^2\le0\)
\(\Rightarrow\dfrac{-\left(x-2\right)^2}{x^2+1}\le0\Rightarrow\dfrac{-\left(x-2\right)^2}{x^2+1}+1\le1\)
vậy GTLN của y là 1\(\Leftrightarrow x-2=0\Leftrightarrow x=2\)
