Đáp án:
$\max A = \dfrac{5}{\sqrt6}\Leftrightarrow x =\dfrac76$
$\min A = \sqrt{\dfrac53}\Leftrightarrow x =\dfrac23$
Giải thích các bước giải:
$A =\sqrt{3 - 2x} + \sqrt{3x-2}\qquad \left(\dfrac23\leq x \leq \dfrac32\right)$
$+)\quad GTLN$
$A =\dfrac{1}{\sqrt3}\sqrt{9-6x} +\dfrac{1}{\sqrt2}\sqrt{6x -4}$
Áp dụng bất đẳng thức $Bunyakovsky$ ta được:
$A^2 \leq \left(\dfrac13+\dfrac12\right)\left(9 - 6x + 6x -5\right)$
$\to A^2 \leq \dfrac{25}{6}$
$\to A \leq \dfrac{5}{\sqrt6}$
Dấu $=$ xảy ra $\Leftrightarrow \sqrt3.\sqrt{9-6x} = \sqrt2.\sqrt{6x -4}$
$\Leftrightarrow x =\dfrac76$
Vậy $\max A = \dfrac{5}{\sqrt6}\Leftrightarrow x =\dfrac76$
$+)\quad GTNN$
$A = \sqrt{3-2x} +\sqrt{\dfrac32}\sqrt{2x - \dfrac43}$
$\to A = \sqrt{3-2x} +\sqrt{2x -\dfrac43} + \dfrac{\sqrt3-\sqrt2}{\sqrt2}\sqrt{2x -\dfrac43}$
Ta có:
$\left(\sqrt{3-2x} +\sqrt{2x -\dfrac43}\right)^2$
$= \dfrac53 + 2\sqrt{(3-2x)\left(2x-\dfrac43\right)}\geq \dfrac53$
$\to \sqrt{3-2x} +\sqrt{2x -\dfrac43} \geq \sqrt{\dfrac53}$
Ta lại có:
$\sqrt{2x -\dfrac43}\geq 0\quad \forall \dfrac23\leq x \leq \dfrac32$
$\to \dfrac{\sqrt3-\sqrt2}{\sqrt2}\sqrt{2x -\dfrac43} \geq 0$
Do đó:
$A = \sqrt{3-2x} +\sqrt{2x -\dfrac43} + \dfrac{\sqrt3-\sqrt2}{\sqrt2}\sqrt{2x -\dfrac43} \geq \sqrt{\dfrac53}$
Dấu $=$ xảy ra $\Leftrightarrow \sqrt{2x -\dfrac43} = 0\Leftrightarrow x =\dfrac23$
Vậy $\min A = \sqrt{\dfrac53}\Leftrightarrow x =\dfrac23$
