Cách khác
\(A=\frac{4x-1}{x^2+3}=\frac{3\left(4x-1\right)}{3\left(x^2+3\right)}=\frac{\left(4x^2+12x+9\right)-4x^2-12}{3\left(x^2+3\right)}=\frac{\left(2x+3\right)^2}{3\left(x^2+3\right)}+\frac{-4\left(x^2+3\right)}{3\left(x^2+3\right)}\)
\(A=\frac{\left(2x+3\right)^2}{3\left(x^2+3\right)}+\frac{-4}{3}\ge-\frac{4}{3}\)
Vậy Min = -4/3 <=> x = -3/2
Đặt \(A=\frac{4x-1}{x^2+3}=t\)\(\Rightarrow x^2.t+3t=4x-1\)
<=> \(x^2.t-4x+3t+1=0\)
Đa thức trên có nghiệm <=> \(\Delta\ge0\)
<=> \(16-4t\left(1+3t\right)\ge0\)
<=> \(16-4t-12t^2\ge0\)
<=> \(3t^2+t-3\le0\)
<=> \(\left(t-1\right)\left(3t+4\right)\le0\)
<=> \(\hept{\begin{cases}t\le1\\t\ge-\frac{4}{3}\end{cases}}\)
Vậy min A = \(-\frac{4}{3}\) <=> \(x=-\frac{3}{2}\)
