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13. 已知$-1＜a＜0$，化简：$\sqrt{a^2 -2+\dfrac{1}{a^2}}-\sqrt{a^2 +2+\dfrac{1}{a^2}}.$
答案
原式$=\sqrt{(a - \dfrac{1}{a})^2} - \sqrt{(a + \dfrac{1}{a})^2}. \because -1 ＜ a ＜ 0, \therefore \dfrac{1}{a} ＜ 0,且\left|\dfrac{1}{a}\right| ＞ |a|. \therefore \sqrt{(a - \dfrac{1}{a})^2} - \sqrt{(a + \dfrac{1}{a})^2} = a - \dfrac{1}{a} - (-a - \dfrac{1}{a}) = a - \dfrac{1}{a} + a + \dfrac{1}{a} = 2a$
14. 如图，认真分析下列各式，然后解答问题.
$(\sqrt{1})^2 +1=2, S_1=\frac{\sqrt{1}}{2}$；
$(\sqrt{2})^2 +1=3, S_2=\frac{\sqrt{2}}{2}$；
$(\sqrt{3})^2 +1=4, S_3=\frac{\sqrt{3}}{2}$；
$\dots$
（1）试用含有$n$（$n$为正整数）的等式表示上述变化规律；
（2）推算出$OA_{10}$的长；
（3）求出$S_1^2 + S_2^2 + S_3^2 + \dots + S_{10}^2$的值.
$(\sqrt{1})^2 +1=2, S_1=\frac{\sqrt{1}}{2}$；
$(\sqrt{2})^2 +1=3, S_2=\frac{\sqrt{2}}{2}$；
$(\sqrt{3})^2 +1=4, S_3=\frac{\sqrt{3}}{2}$；
$\dots$
（1）试用含有$n$（$n$为正整数）的等式表示上述变化规律；
（2）推算出$OA_{10}$的长；
（3）求出$S_1^2 + S_2^2 + S_3^2 + \dots + S_{10}^2$的值.
答案
(1) 由图形知，$OA_1^2 + A_1A_2^2 = OA_2^2$，取$(\sqrt{1})^2 + 1 = 2$，$S_1=\dfrac{1}{2}OA_1 · A_1A_2=\dfrac{\sqrt{1}}{2}$. 以此类推，可知$(\sqrt{n})^2 + 1 = n + 1$，$S_n=\dfrac{\sqrt{n}}{2}$
(2) $\because OA_1=1$，$OA_2=\sqrt{2}$，$OA_3=\sqrt{3}$，$\dots$，$\therefore OA_{10}=\sqrt{10}$
(3) $S_1^2 + S_2^2 + \dots + S_{10}^2 = (\dfrac{\sqrt{1}}{2})^2 + (\dfrac{\sqrt{2}}{2})^2 + \dots + (\dfrac{\sqrt{10}}{2})^2 = \dfrac{1}{4} + \dfrac{2}{4} + \dots + \dfrac{10}{4} = \dfrac{1}{4}(1 + 2 + \dots + 10) = \dfrac{55}{4}$
(2) $\because OA_1=1$，$OA_2=\sqrt{2}$，$OA_3=\sqrt{3}$，$\dots$，$\therefore OA_{10}=\sqrt{10}$
(3) $S_1^2 + S_2^2 + \dots + S_{10}^2 = (\dfrac{\sqrt{1}}{2})^2 + (\dfrac{\sqrt{2}}{2})^2 + \dots + (\dfrac{\sqrt{10}}{2})^2 = \dfrac{1}{4} + \dfrac{2}{4} + \dots + \dfrac{10}{4} = \dfrac{1}{4}(1 + 2 + \dots + 10) = \dfrac{55}{4}$
