Ta có
\(\frac{4x+3}{x^2+1}=\frac{-\left(x^2+1\right)+x^2+4x+4}{x^2+1}=-1+\frac{\left(x+2\right)^2}{x^2+1}\ge-1\)
Dấu ''='' xảy ra <=>x=-2
Ta có
\(\frac{4x+3}{x^2+1}=\frac{4\left(x^2+1\right)-4x^2+4x-1}{x^2+1}=4-\frac{\left(2x-1\right)^2}{x^2+1}\le4\)
Dấu ''='' xảy ra <=>x=1/2