Đáp án:
Bài 1:
c) $\sqrt{\dfrac{2+\sqrt{3}}{2}}-\dfrac{\sqrt{3}}{2}=\dfrac{1}{2}$
Bài 2:
c) $x=0$
Giải thích các bước giải:
Bài 1:
c)
$\sqrt{\dfrac{2+\sqrt{3}}{2}}-\dfrac{\sqrt{3}}{2}$
$=\sqrt{\dfrac{4+2\sqrt{3}}{4}}-\dfrac{\sqrt{3}}{2}$
$=\sqrt{\dfrac{(\sqrt{3}+1)^2}{4}}-\dfrac{\sqrt{3}}{2}$
$=\dfrac{|\sqrt{3}+1|}{2}-\dfrac{\sqrt{3}}{2}$
$=\dfrac{\sqrt{3}+1-\sqrt{3}}{2}$
$=\dfrac{1}{2}$
Bài 2:
c)
$A=\dfrac{\sqrt{x}+2}{\sqrt{x}+1}\,\,(x \ge 0)$
$=\dfrac{\sqrt{x}+1+1}{\sqrt{x}+1}$
$=1+\dfrac{1}{\sqrt{x}+1}$
Để A nguyên $\to 1 \vdots (\sqrt{x}+1)$
$\to (\sqrt{x}+1) \in Ư(1)=\{\pm1\}$
$\to \left[\begin{array}{l}\sqrt{x}+1=1\\\sqrt{x}+1=-1\end{array}\right.$
$\to \left[\begin{array}{l}\sqrt{x}=0\\\sqrt{x}=-2\text{ (loại)}\end{array}\right.$
$\to x=0$ (thỏa mãn)
Vậy $x=0$ thì A nguyên
