Đáp án:
$\min\left(4a + \dfrac{9}{a - 3} + 2020\right)= 2020 \Leftrightarrow a = \dfrac92$
Giải thích các bước giải:
Đặt $P = 4a + \dfrac{9}{a - 3} + 2020$
$\to P = 4a - 12 + \dfrac{9}{a-3} + 2008$
$\to P = 4(a-3) + \dfrac{9}{a-3} + 2008$
$\to P \geqslant 2\sqrt{4(a-3)\cdot\dfrac{9}{a-3}} + 2008\quad (BDT\ AM-GM)$
$\to P \geqslant 2\cdot 6 + 2008$
$\to P \geqslant 2020$
Dấu $=$ xảy ra $\Leftrightarrow 4(a-3) =\dfrac{9}{a-3}$
$\Leftrightarrow (a-3)^2 =\dfrac94$
$\Leftrightarrow a = \dfrac92$
Vậy $\min\left(4a + \dfrac{9}{a - 3} + 2020\right)= 2020 \Leftrightarrow a = \dfrac92$
