Tìm $GTNN$ và $GTLN$ của `A=\sqrt(3-2x)+\sqrt(3x-2)`

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 $\min A = \sqrt{\dfrac53}\Leftrightarrow x =\dfrac23$