Ta có: \(\frac{3x^2-8x+6}{x^2-2x+1}=B\)
=>\(B=\frac{3x^2-6x+3-2x+3}{x^2-2x+1}=3+\frac{-2x+3}{\left(x-1\right)^2}\)
\(=3+\frac{-2x+2+1}{\left(x-1\right)^2}=3-\frac{2}{\left(x-1\right)}+\frac{1}{\left(x-1\right)^2}\)
\(=\left(\frac{1}{x-1}-1\right)^2+2\ge2\forall x\)
Dấu '=' xảy ra khi \(\frac{1}{x-1}-1=0\)
=>x-1=1
=>x=2