Đáp án:
a) $max A = \dfrac{4}{3} \Leftrightarrow x = \dfrac{1}{4}$
b) $maxB = \dfrac{1}{3} \Leftrightarrow x = 1$
c) $maxC= 1 \Leftrightarrow x = 1$
Giải thích các bước giải:
a) $A = \dfrac{1}{x - \sqrt x +1}$
$= \dfrac{1}{x - 2.\dfrac{1}{2}\sqrt x + \dfrac{1}{4} + \dfrac{3}{4}}$
$= \dfrac{1}{\left(\sqrt x -\dfrac{1}{2}\right)^2 + \dfrac{3}{4}}$
Ta có: $\left(\sqrt x -\dfrac{1}{2}\right)^2 \geq 0, \forall x$
$\Rightarrow \left(\sqrt x -\dfrac{1}{2}\right)^2 + \dfrac{3}{4} \geq \dfrac{3}{4}$
$\Rightarrow \dfrac{1}{\left(\sqrt x -\dfrac{1}{2}\right)^2 + \dfrac{3}{4}} \leq \dfrac{4}{3}$
Hay $A \leq \dfrac{4}{3}$
Dấu = xảy ra $\Leftrightarrow \sqrt x - \dfrac{1}{2} = 0 \Leftrightarrow x = \dfrac{1}{4}$
Vậy $max A = \dfrac{4}{3} \Leftrightarrow x = \dfrac{1}{4}$
b) $B = \dfrac{\sqrt x}{x + \sqrt x +1}$
$= \dfrac{1}{\sqrt x + 1 + \dfrac{1}{\sqrt x}}$
Ta có: $\sqrt x + \dfrac{1}{\sqrt x} \geq 2\sqrt{\sqrt x.\dfrac{1}{\sqrt x}} = 2$
$\Leftrightarrow \sqrt x + 1 + \dfrac{1}{\sqrt x} \geq 3$
$\Leftrightarrow \dfrac{1}{\sqrt x + 1 + \dfrac{1}{\sqrt x}} \leq \dfrac{1}{3}$
Hay $B \leq 3$
Dấu = xảy ra $\Leftrightarrow \sqrt x = \dfrac{1}{\sqrt x} \Leftrightarrow x = 1$
Vậy $maxB = \dfrac{1}{3} \Leftrightarrow x = 1$
c) $C = \dfrac{\sqrt x}{x - \sqrt x +1}$
$= \dfrac{1}{\sqrt x - 1 + \dfrac{1}{\sqrt x}}$
Ta có: $\sqrt x + \dfrac{1}{\sqrt x} \geq 2\sqrt{\sqrt x.\dfrac{1}{\sqrt x}} = 2$
$\Leftrightarrow \sqrt x - 1 + \dfrac{1}{\sqrt x} \geq 1$
$\Leftrightarrow \dfrac{1}{\sqrt x - 1 + \dfrac{1}{\sqrt x}} \leq 1$
Hay $C \leq 1$
Dấu = xảy ra $\Leftrightarrow \sqrt x = \dfrac{1}{\sqrt x} \Leftrightarrow x = 1$
Vậy $maxC= 1 \Leftrightarrow x = 1$
