《2025年名校题库八年级数学上册北师大版》

16. (高新区期末)已知$x= \sqrt {3}-1$,$y= \sqrt {3}+1$.
(1)求$x^{2}+xy+y^{2}$的值；
(2)若$a是x$的小数部分,$b是y$的整数部分,求$\frac {1}{a+b}-\frac {b}{a}$的值.

答案： 解:
(1) 原式 $= (x + y)^2 - xy = (2\sqrt{3})^2 - 2 = 12 - 2 = 10$.
(2) $a = \sqrt{3} - 1$, $b = 2$, 原式 $= \frac{1}{\sqrt{3} - 1 + 2} - \frac{2}{\sqrt{3} - 1} = \frac{1}{\sqrt{3} + 1} - \frac{2}{\sqrt{3} - 1} = \frac{\sqrt{3} - 1}{2} - (\sqrt{3} + 1) = \frac{\sqrt{3}}{2} - \frac{1}{2} - \sqrt{3} - 1 = -\frac{\sqrt{3}}{2} - \frac{3}{2}$.

17. (树德实验)已知$x= \frac {1}{\sqrt {5}-2}$,$y= \frac {1}{\sqrt {5}+2}$.
(1)求$x^{2}+xy+y^{2}$的值；
(2)若$x的小数部分为a$,$y的整数部分为b$,求$ax+by$的平方根.

答案： 解:
(1) $x = \sqrt{5} + 2$, $y = \sqrt{5} - 2$, 原式 $= (x + y)^2 - xy = (\sqrt{5} + 2 + \sqrt{5} - 2)^2 - (\sqrt{5} + 2)(\sqrt{5} - 2) = 20 - 1 = 19$.
(2) $a = \sqrt{5} - 2$, $b = 0$, 原式 $= (\sqrt{5} - 2)(\sqrt{5} + 2) + 0 = 1$, $\therefore ax + by$ 的平方根是 $\pm 1$.

18. (成外)
(1)观察下列各式的特点：$\sqrt {2}-1＞\sqrt {3}-\sqrt {2}$,$\sqrt {3}-\sqrt {2}＞\sqrt {4}-\sqrt {3}$,$\sqrt {4}-\sqrt {3}＞\sqrt {5}-\sqrt {4}$,…根据以上规律可知：$\sqrt {2018}-\sqrt {2017}$＜$\sqrt {2017}-\sqrt {2016}$.
(2)观察下列式子的化简过程：$\frac {1}{\sqrt {2}+1}= \frac {\sqrt {2}-1}{(\sqrt {2}+1)(\sqrt {2}-1)}= \sqrt {2}-1$,$\frac {1}{\sqrt {3}+\sqrt {2}}= \frac {\sqrt {3}-\sqrt {2}}{(\sqrt {3}+\sqrt {2})(\sqrt {3}-\sqrt {2})}= \sqrt {3}-\sqrt {2}$,$\frac {1}{\sqrt {4}+\sqrt {3}}= \frac {\sqrt {4}-\sqrt {3}}{(\sqrt {4}+\sqrt {3})(\sqrt {4}-\sqrt {3})}= \sqrt {4}-\sqrt {3}$,…根据观察,请写出式子$\frac {1}{\sqrt {n+1}-\sqrt {n}}(n≥1)$的化简过程.
解: $\frac{1}{\sqrt{n + 1} - \sqrt{n}} = \frac{\sqrt{n + 1} + \sqrt{n}}{(\sqrt{n + 1} - \sqrt{n})(\sqrt{n + 1} + \sqrt{n})} = \sqrt{n + 1} + \sqrt{n}$.
(3)计算下列算式：$\frac {1}{\sqrt {3}+1}+\frac {1}{\sqrt {5}+\sqrt {3}}+\frac {1}{\sqrt {7}+\sqrt {5}}+... +\frac {1}{\sqrt {2019}+\sqrt {2017}}$.
解: 原式 $= \frac{\sqrt{3} - 1}{2} + \frac{\sqrt{5} - \sqrt{3}}{2} + \frac{\sqrt{7} - \sqrt{5}}{2} + \cdots + \frac{\sqrt{2019} - \sqrt{2017}}{2} = -\frac{1}{2} + \frac{\sqrt{3}}{2} - \frac{\sqrt{3}}{2} + \frac{\sqrt{5}}{2} - \frac{\sqrt{5}}{2} + \frac{\sqrt{7}}{2} + \cdots - \frac{\sqrt{2017}}{2} + \frac{\sqrt{2019}}{2} = -\frac{1}{2} + \frac{\sqrt{2019}}{2} = \frac{\sqrt{2019} - 1}{2}$.

答案：
(1) $＜$
(2) 解: $\frac{1}{\sqrt{n + 1} - \sqrt{n}} = \frac{\sqrt{n + 1} + \sqrt{n}}{(\sqrt{n + 1} - \sqrt{n})(\sqrt{n + 1} + \sqrt{n})} = \sqrt{n + 1} + \sqrt{n}$.
(3) 解: 原式 $= \frac{\sqrt{3} - 1}{2} + \frac{\sqrt{5} - \sqrt{3}}{2} + \frac{\sqrt{7} - \sqrt{5}}{2} + \cdots + \frac{\sqrt{2019} - \sqrt{2017}}{2} = -\frac{1}{2} + \frac{\sqrt{3}}{2} - \frac{\sqrt{3}}{2} + \frac{\sqrt{5}}{2} - \frac{\sqrt{5}}{2} + \frac{\sqrt{7}}{2} + \cdots - \frac{\sqrt{2017}}{2} + \frac{\sqrt{2019}}{2} = -\frac{1}{2} + \frac{\sqrt{2019}}{2} = \frac{\sqrt{2019} - 1}{2}$.