Đáp án:
$\min B = 2\sqrt3 +\dfrac32\Leftrightarrow x = 1 +\dfrac{2}{\sqrt3}$
Giải thích các bước giải:
$B =\dfrac{3x}{2} +\dfrac{2}{x-1}\qquad (x > 1)$
$\to B =\dfrac{3(x-1) + 3}{2} +\dfrac{2}{x-1}$
$\to B =\dfrac{3(x-1)}{2} +\dfrac{2}{x-1} +\dfrac{3}{2}$
Áp dụng bất đẳng thức $AM-GM$ ta được:
$\dfrac{3(x-1)}{2} +\dfrac{2}{x-1}\geq 2\sqrt{\dfrac{3(x-1)}{2}\cdot\dfrac{2}{x-1}}$
$\to \dfrac{3(x-1)}{2} +\dfrac{2}{x-1}\geq 2\sqrt3$
$\to \dfrac{3(x-1)}{2} +\dfrac{2}{x-1} +\dfrac{3}{2}\geq 2\sqrt3 +\dfrac32$
$\to B \geq 2\sqrt3 +\dfrac32$
Dấu $=$ xảy ra $\Leftrightarrow \dfrac{3(x-1)}{2} =\dfrac{2}{x-1}$
$\Leftrightarrow (x-1)^2 =\dfrac43$
$\Leftrightarrow x = 1 +\dfrac{2}{\sqrt3}\quad (Do\,\,x > 1)$
Vậy $\min B = 2\sqrt3 +\dfrac32\Leftrightarrow x = 1 +\dfrac{2}{\sqrt3}$
