Lời giải:
\(B=\frac{x^2-3x+3}{x^2-2x+1}\)
\(B=\frac{(x^2-2x+1)-x+1+1}{\left(x-1\right)^2}\)
\(B=\frac{(x-1)^2-\left(x-1\right)+1}{\left(x-1\right)^2}\)
\(B=\frac{(x-1)^2}{\left(x-1\right)^2}-\frac{x-1}{\left(x-1\right)^2}+\frac{1}{\left(x-1\right)^2}\)
\(B=1-\frac{1}{x-1}+\frac{1}{\left(x-1\right)^2}\)
Đặt: \(\frac{1}{x-1}=a\), ta có:
\(B=a^2-a+1\)
\(B=a^2-2.a.\frac{1}{2}+\left(\frac{1}{2}\right)^2-\left(\frac{1}{2}\right)^2+1\)
\(B=\left(a-\frac{1}{2}\right)+\frac{3}{4}\ge\frac{3}{4}\)∀x
\(\Rightarrow B_{min}=\frac{3}{4}\Leftrightarrow\left(a-\frac{1}{2}\right)^2=0\Leftrightarrow a=\frac{1}{2}\Leftrightarrow x=3\)
Vậy:\(B_{min}=\frac{3}{4}\Leftrightarrow x=3\)
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