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10.在1，-2，$-\sqrt{3}$，0，π五个数中，最小的数是
-2
。答案
10.-2
11. 如图所示,直径为1个单位长度的圆从原点沿数轴向右滚动一周(不滑动),圆上的一点由原点到达点$O'$,点$O'$的数值是
π
.答案
11.π
12. 若将三个数$-\sqrt{3}$，$\sqrt{7}$，$\sqrt{11}$表示在数轴上，其中能被如图所示的墨迹覆盖的数是
$\sqrt{7}$
。答案
12.$\sqrt{7}$
三、解答题
13. 计算：
(1) $\sqrt{3^{-2}} + \sqrt[3]{-\dfrac{64}{27}}$；
(2) $\sqrt[3]{0.027} - \sqrt[3]{-8} - |1 - \sqrt{3}|$；
13. 计算：
(1) $\sqrt{3^{-2}} + \sqrt[3]{-\dfrac{64}{27}}$；
(2) $\sqrt[3]{0.027} - \sqrt[3]{-8} - |1 - \sqrt{3}|$；
答案
13.(1) -1 (2)$3.3 - \sqrt{3}$
14. 求下列各式中$x$的值：
(1)$49x^2 -16 = 0$；
(2)$8x^3 +125 = 0$。
(1)$49x^2 -16 = 0$；
(2)$8x^3 +125 = 0$。
答案
14.解:(1)$x = \pm \dfrac{4}{7}$;(2)$x = -\dfrac{5}{2}$.
15.已知$|a|=2-\sqrt{2}$,$|b|=3-2\sqrt{2}$,又$a+b=\sqrt{2}-1$,求a和b的值.
答案
15.解:由$|a| =2 -\sqrt{2}$得$a = \pm (2 -\sqrt{2})$;
由$|b| =3 -2\sqrt{2}$得$b = \pm (3 -2\sqrt{2})$.
当$a =2 -\sqrt{2}$,$b =3 -2\sqrt{2}$时,$a + b =5 -3\sqrt{2}$;
当$a =2 -\sqrt{2}$,$b = -(3 -2\sqrt{2})$时,$a + b =\sqrt{2} -1$;
当$a = -(2 -\sqrt{2})$,$b =3 -2\sqrt{2}$时,$a + b =1 -\sqrt{2}$;
当$a = -(2 -\sqrt{2})$,$b = -(3 -2\sqrt{2})$时,
$a + b = -5 +3\sqrt{2}$. 又$a + b =\sqrt{2} -1$,
因此$a =2 -\sqrt{2}$,$b =2\sqrt{2} -3$.
由$|b| =3 -2\sqrt{2}$得$b = \pm (3 -2\sqrt{2})$.
当$a =2 -\sqrt{2}$,$b =3 -2\sqrt{2}$时,$a + b =5 -3\sqrt{2}$;
当$a =2 -\sqrt{2}$,$b = -(3 -2\sqrt{2})$时,$a + b =\sqrt{2} -1$;
当$a = -(2 -\sqrt{2})$,$b =3 -2\sqrt{2}$时,$a + b =1 -\sqrt{2}$;
当$a = -(2 -\sqrt{2})$,$b = -(3 -2\sqrt{2})$时,
$a + b = -5 +3\sqrt{2}$. 又$a + b =\sqrt{2} -1$,
因此$a =2 -\sqrt{2}$,$b =2\sqrt{2} -3$.
