+) Tìm Min
\(A=\frac{4x+3}{x^2+1}=>A+1=\frac{x^2+4x+4}{x^2+1}=\frac{\left(x+2\right)^2}{x^2+1}>=0\)
\(=>MinA=-1\)
Dấu "=" xảy ra khi x+2=0<=>x=-2
+)tìm Max
\(A=\frac{4x+3}{x^2+1}=4-\frac{4x^2-4x+1}{x^2+1}=4-\frac{\left(2x-1\right)^2}{x^2+1}< =4\)
=>Max A=4
Dấu "=" xảy ra khi 2x-1=0<=>x=1/2