a. Ta có: \(Q=3x^2-4x+7=3\left(x^2-\frac{4}{3}x+\frac{7}{3}\right)\\ =3\left[\left(x^2-2.x.\frac{2}{3}+\frac{4}{9}\right)+\frac{17}{9}\right]\\ =3\left[\left(x-\frac{2}{3}\right)^2+\frac{17}{9}\right]\\ =3\left(x-\frac{2}{3}\right)^2+\frac{17}{3}\)
Mà \(\left(x-\frac{2}{3}\right)^2\ge0\forall x\Rightarrow3\left(x-\frac{2}{3}\right)^2\ge0\forall x\\ \Rightarrow3\left(x-\frac{2}{3}\right)^2+\frac{17}{3}\ge\frac{17}{3}>0\forall x\)
Hay \(Q>0\forall x\)
b. Theo câu a ta có:
Q = \(3\left(x-\frac{2}{3}\right)^2+\frac{17}{3}\ge\frac{17}{3}\forall x\)
Dấu = xảy ra khi: \(3\left(x-\frac{2}{3}\right)^2=0\\ \Leftrightarrow\left(x-\frac{2}{3}\right)^2=0\\ \Leftrightarrow x-\frac{2}{3}=0\Leftrightarrow x=\frac{2}{3}\)
Vậy \(MinQ=\frac{17}{3}\Leftrightarrow x=\frac{2}{3}\)
Có: \(Q=3\left(x-\frac{2}{3}\right)^2+\frac{17}{3}\ge\frac{17}{3}>0\forall x\in R\)(cái này bạn nhóm lại là ra)
Bài toán đã đc giải quyết.
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