Đáp án:
$\min N = \dfrac{23}{12} \Leftrightarrow x = \dfrac{1}{6}$
Giải thích các bước giải:
$N = (3x + 2)(x - 1) +4$
$= 3x^2 - x + 2$
$= 3\left(x^2 - \dfrac{1}{3}x + \dfrac{2}{3}\right)$
$= 3\left(x^2 - 2.\dfrac{1}{6}.x + \dfrac{1}{36}\right) + \dfrac{23}{12}$
$ = 3\left(x - \dfrac{1}{6}\right)^2 + \dfrac{23}{12}$
Ta có: $3\left(x - \dfrac{1}{6}\right)^2 \geq 0, \,\forall x$
$\Leftrightarrow 3\left(x - \dfrac{1}{6}\right)^2 + \dfrac{23}{12} \geq \dfrac{23}{12}$
Hay $N \geq \dfrac{23}{12}$
Dấu = xảy ra $\Leftrightarrow x = \dfrac{1}{6}$
Vậy $\min N = \dfrac{23}{12} \Leftrightarrow x = \dfrac{1}{6}$
