Đáp án:
Giải thích các bước giải:
a.
$x \leq \dfrac{1}{2} ⇒\dfrac{1}{x} \geq 2$
$A=x+\dfrac{1}{x}=x+\dfrac{1}{4x}+\dfrac{3}{4}.\dfrac{1}{x}$
$A \geq 2\sqrt{x·\dfrac{1}{4x}}+\dfrac{3}{4}·2=\dfrac{5}{2}$
$A_{min}=\dfrac{5}{2}$ khi $x=\dfrac{1}{2}$
b.
$A=x+\dfrac{1}{x}=\dfrac{x}{4}+\dfrac{1}{x}+\dfrac{3}{4}·x$
$A \geq 2\sqrt{\dfrac{x}{4}·\dfrac{1}{x}}+\dfrac{3}{4}·2=\dfrac{5}{2}$
$A_{min}=\dfrac{5}{2}$ khi $x=2$
