$A = 5x^2 - 3x + 7$
$\to 5A = 25x^2 - 15x + 35$
$\to 5A = (5x)^2 - 2.5x.\dfrac{3}{2} + \dfrac{9}{4} - \dfrac{9}{4} + 35$
$\to 5A = ( 5x - \dfrac{3}{2})^2 + \dfrac{131}{4}$
$\to A = \dfrac{ ( 5x - \dfrac{3}{2})^2}{5} + \dfrac{131}{20}$
Ta có: $( 5x - \dfrac{3}{2})^2 \geq 0$ $\forall \, \, \, x \, \, \in \, \, \mathbb{R}$
$\to \dfrac{( 5x - \dfrac{3}{2})^2}{5} \geq 0 $ $\forall \, \, \, x \, \, \in \, \, \mathbb{R}$
$\to \dfrac{( 5x - \dfrac{3}{2})^2}{5} + \dfrac{131}{20} \geq \dfrac{131}{20} $ $\forall \, \, \, x \, \, \in \, \, \mathbb{R}$
Dấu "$=$" xảy ra khi và chỉ khi $ \dfrac{( 5x - \dfrac{3}{2})^2}{5} = 0$
$\Leftrightarrow ( 5x - \dfrac{3}{2})^2 = 0$
$\Leftrightarrow 5x - \dfrac{3}{2} = 0$
$\Leftrightarrow x = 0,3$
Vậy $Min_A$ = $\dfrac{131}{20}$ khi và chỉ khi $x = 0,3$
