Đặt A =\(\frac{3-4x}{x^2+1}\)
Xét A + 1 = \(\frac{3-4x}{x^2+1}+1\)
\(=\frac{3-4x}{x^2+1}+\frac{x^2+1}{x^2+1}\)
\(=\frac{x^2-4x+4}{x^2+1}\)
\(=\frac{\left(x-2\right)^2}{x^2+1}\)
Có (x - 2)2 \(\ge\)0
x2 + 1 > 0
=> \(\frac{\left(x-2\right)^2}{x^2+1}\ge0\)
=> A + 1 \(\ge\)0
=> A \(\ge\)-1
Dấu "=" xảy ra <=> (x - 2)2 = 0 <=> x - 2 = 0 <=> x = 2
Vậy Amin = -1 <=> x = 2