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8. (2024·上海)
计算:
$|1-\sqrt {3}|+\sqrt {24}+\frac {1}{2+\sqrt {3}}-(1-\sqrt {3})^{0}.$
计算:
$|1-\sqrt {3}|+\sqrt {24}+\frac {1}{2+\sqrt {3}}-(1-\sqrt {3})^{0}.$
答案
$ 8. 2 \sqrt { 6 } $
二、二次根式的性质
(1) $\sqrt{a}$0($a≥0$).
(2) $(\sqrt{a})^2$ = ($a≥0$).
(3) $\sqrt{a^2}$ = ______ = $\begin{cases}\_\_\_\_\_\_, \\ \_\_\_\_\_\_, \\ \_\_\_\_\_\_.\end{cases}$
(4) $\sqrt{ab}$ = ($a≥0$, $b≥0$).
(5) $\sqrt{\dfrac{a}{b}}$ = ($a≥0$, $b＞0$).
(1) $\sqrt{a}$0($a≥0$).
(2) $(\sqrt{a})^2$ = ($a≥0$).
(3) $\sqrt{a^2}$ = ______ = $\begin{cases}\_\_\_\_\_\_, \\ \_\_\_\_\_\_, \\ \_\_\_\_\_\_.\end{cases}$
(4) $\sqrt{ab}$ = ($a≥0$, $b≥0$).
(5) $\sqrt{\dfrac{a}{b}}$ = ($a≥0$, $b＞0$).
答案
(1) ≥
(2) a
(3) |a|；a(a＞0)；0(a=0)；-a(a＜0)
(4) $\sqrt{a}·\sqrt{b}$
(5) $\dfrac{\sqrt{a}}{\sqrt{b}}$
(2) a
(3) |a|；a(a＞0)；0(a=0)；-a(a＜0)
(4) $\sqrt{a}·\sqrt{b}$
(5) $\dfrac{\sqrt{a}}{\sqrt{b}}$
