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2. 计算：
(1)$(1-\sqrt {3})^{0}+|3-\sqrt {12}|-\frac {6}{\sqrt {3}}$
(2)$\sqrt {24}×\sqrt {\frac {1}{3}}-4×\sqrt {\frac {1}{8}}×(1-\sqrt {2})^{0}$
(1)$(1-\sqrt {3})^{0}+|3-\sqrt {12}|-\frac {6}{\sqrt {3}}$
(2)$\sqrt {24}×\sqrt {\frac {1}{3}}-4×\sqrt {\frac {1}{8}}×(1-\sqrt {2})^{0}$
答案
(1) -2 (2) $\sqrt{2}$
3. 先化简,再求值：$(\frac {a^{2}-b^{2}}{a^{2}-2ab+b^{2}}+\frac {a}{b-a})÷\frac {b^{2}}{a^{2}-ab}$,其中$a$,$b满足\sqrt {a+1}+|b-\sqrt {3}|= 0$。
答案
$-\frac{\sqrt{3}}{3}$
