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1.计算：$|-2|+\sqrt{4}-(-1)^2$
答案
解：$|-2|+\sqrt{4}-(-1)^2=2+2-1=3.$
2. 化简：$\sqrt{18} - \sqrt{\frac{9}{2}} - \frac{\sqrt{3} + \sqrt{6}}{\sqrt{3}} + (\sqrt{3} - 2)^0 + \sqrt{(1 - \sqrt{2})^2}$
答案
解:原式$=3\sqrt{2}-\frac{3}{2}\sqrt{2}-(1+\sqrt{2})+1+|1-\sqrt{2}|$
$=3\sqrt{2}-\frac{3}{2}\sqrt{2}-1-\sqrt{2}+1+\sqrt{2}-1$
$=\frac{3}{2}\sqrt{2}-1.$
$=3\sqrt{2}-\frac{3}{2}\sqrt{2}-1-\sqrt{2}+1+\sqrt{2}-1$
$=\frac{3}{2}\sqrt{2}-1.$
3.解不等式组：$\begin{cases} x+3＞0①, \\ 2(x-1)+3≥ 3x② \end{cases}$ 并判断 $x=\dfrac{\sqrt{3}}{2}$ 是否满足该不等式组.
答案
解:由①得:$x ＞ -3$.
由②得:$x ≤ 1$.
$\therefore$原不等式组的解集是:$-3 ＜ x ≤ 1$.
$\therefore x = \frac{\sqrt{3}}{2}$满足该不等式组.
由②得:$x ≤ 1$.
$\therefore$原不等式组的解集是:$-3 ＜ x ≤ 1$.
$\therefore x = \frac{\sqrt{3}}{2}$满足该不等式组.
