\(D=\frac{4x+3}{x^2+1}\)
Min D :
\(D=\frac{x^2+4x+4-x^2-1}{x^2+1}\)
\(=\frac{\left(x+2\right)^2-\left(x^2+1\right)}{x^2+1}=\frac{\left(x+2\right)^2}{x^2+1}-1\)
Ta thấy : \(\frac{\left(x+2\right)^2}{x^2+1}\ge0\forall x\)
\(\Rightarrow D\Rightarrow\frac{\left(x+2\right)^2}{x^2+1}-1\ge-1\)
Dấu "=" xảy ra khi \(x+2=0\Leftrightarrow x=-2\)
Max D :
\(D=\frac{4x+3}{x^2+1}=\frac{-4x^2+4x-1+4x^2+4}{x^2+1}\)
\(=\frac{-\left(2x-1\right)^2+4\left(x^2+1\right)}{x^2+1}\)
\(=\frac{-\left(2x-1\right)^2}{x^2+1}+4\)
Ta thấy : \(\frac{-\left(2x-1\right)^2}{x^2+1}\le0\forall x\)
\(\Rightarrow D=\frac{-\left(2x-1\right)^2}{x^2+1}+4\le4\)
Dấu "=" xảy ra khi \(2x-1=0\Leftrightarrow x=\frac{1}{2}\)
