ĐKXĐ x ≠1
\(B=\dfrac{3x^2-8x+6}{x^2-2x+1}\)
= \(\dfrac{\left(2x^2-4x+2\right)+\left(x^2-4x+4\right)}{x^2-2x+1}\)
= \(\dfrac{2\left(x^2-2x+1\right)}{x^2-2x+1}+\dfrac{x^2-4x+4}{x^2-2x+1}\)
= \(2+\dfrac{\left(x-2\right)^2}{\left(x-1\right)^2}\)
do (x-2)2 ≥0 ∀x
(x-1)2 ≥0 ∀x
=> \(\dfrac{\left(x-2\right)^2}{\left(x-1\right)^2}\ge0\)
<=> \(2+\dfrac{\left(x-2\right)^2}{\left(x-1\right)^2}\ge2\)
<=> B ≥ 2
Min B =2 khi
(x-2)2 =0
⇔x-2=0
⇔x=2
vậy GTNN B =2 khi x=2
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