Sửa đề: Tìm GTNN....
\(h\left(x\right)=3x^2-4x+3=3\left(x^2-\dfrac{4}{3}x+1\right)\)
\(=3\left[\left(x^2-2\cdot x\cdot\dfrac{2}{3}+\dfrac{4}{9}\right)+\dfrac{5}{9}\right]\)
\(=3\left[\left(x-\dfrac{2}{3}\right)^2+\dfrac{5}{9}\right]\)
Vì \(\left(x-\dfrac{2}{3}\right)^2+\dfrac{5}{9}\ge\dfrac{5}{9}\)
=> \(3\left[\left(x-\dfrac{2}{3}\right)^2+\dfrac{5}{9}\right]\ge3\cdot\dfrac{5}{9}=\dfrac{15}{9}\)
Dấu ''='' xảy ra khi \(x=\dfrac{2}{3}\)
Vậy GTNN của h(x) là \(\dfrac{15}{9}\Leftrightarrow x=\dfrac{2}{3}\)
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