\(B=\frac{x^2-3x+3}{x^2-2x+1}\)
\(B=\frac{x^2-2x+1-x+1+1}{x^2-2x+1}\)
\(B=\frac{\left(x-1\right)^2-\left(x-1\right)+1}{\left(x-1\right)^2}\)
\(B=1-\frac{1}{x-1}+\frac{1}{\left(x-1\right)^2}\)
Đặt \(\frac{1}{x-1}=a\)
\(B=a^2-a+1\)
\(B=a^2-2\cdot a\cdot\frac{1}{2}+\frac{1}{4}+\frac{3}{4}\)
\(B=\left(a-\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}\forall a\)
Dấu "=" xảy ra \(\Leftrightarrow a=\frac{1}{2}\Leftrightarrow\frac{1}{x-1}=\frac{1}{2}\Leftrightarrow x=3\)
Lời giải:
Ta có: \(B=\frac{x^2-3x+3}{x^2-2x+1}=\frac{x^2-3x+3}{\left(x-1\right)^2}\)
Đặt: \(x-1=a\Rightarrow x=a+1\), ta có:
\(B=\frac{\left(a+1\right)^2-3\left(a+1\right)+3^2}{a^2}=\frac{a^2+2a+1-3a-3+9}{a^2}=\frac{a^2-a+7}{a^2}=\frac{a^2}{a^2}-\frac{a}{a^2}+\frac{7}{a^2}=1-\frac{1}{a}+\frac{7}{a^2}=\frac{7}{a^2}-\frac{1}{a}+1\)
Đặt: \(\frac{1}{a}=b\), ta có:
\(B=7b^2-b+1\)
\(\Rightarrow\frac{B}{7}=b^2-\frac{1}{7}b+\frac{1}{7}\)
\(=b^2-2.\frac{1}{14}b+\left(\frac{1}{14}\right)^2-\left(\frac{1}{14}\right)^2+\frac{1}{7}\)
\(=\left(b-\frac{1}{14}\right)^2-\frac{27}{196}\)
\(\Rightarrow B=7\left(b-\frac{1}{14}\right)^2-\frac{27}{28}\ge-\frac{27}{28}\).Vì: \(7\left(b-\frac{1}{14}\right)^2\ge0\)∀x
\(\Rightarrow B_{min}=-\frac{27}{28}\Leftrightarrow7\left(b-\frac{1}{14}\right)^2=0\Leftrightarrow\left(b-\frac{1}{14}\right)^2=0\Leftrightarrow b=\frac{1}{14}\Leftrightarrow a=14\Leftrightarrow x=15\)
Vậy:\(\Rightarrow B_{min}=-\frac{27}{28}\Leftrightarrow x=15\)
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