《2025年名校题库八年级数学上册北师大版》

12. (七中育才)化简求值：
(1)已知$x= \frac {1}{\sqrt {3}-\sqrt {2}}$,$y= \frac {1}{\sqrt {3}+\sqrt {2}}$,求$x^{2}-xy+y^{2}$的值；
(2)已知$x= \frac {3+\sqrt {5}}{2}$,$y= \frac {3-\sqrt {5}}{2}$,求$x^{2}-xy+y^{2}$的值.

答案： 解:
(1) $x = \frac{1}{\sqrt{3} - \sqrt{2}} = \sqrt{3} + \sqrt{2}$, $y = \frac{1}{\sqrt{3} + \sqrt{2}} = \sqrt{3} - \sqrt{2}$.
原式 $= (x - y)^2 + xy = (\sqrt{3} + \sqrt{2} - \sqrt{3} + \sqrt{2})^2 + (\sqrt{3} + \sqrt{2})(\sqrt{3} - \sqrt{2}) = 8 + 1 = 9$.
(2) 原式 $= (x - y)^2 + xy = (\frac{3 + \sqrt{5}}{2} - \frac{3 - \sqrt{5}}{2})^2 + \frac{3 + \sqrt{5}}{2} \cdot \frac{3 - \sqrt{5}}{2} = 5 + 1 = 6$.

13. (锦江区期末)已知$a= \frac {1}{\sqrt {2}-1}$,$b= \frac {1}{\sqrt {2}+1}$,求$a^{2}-3ab+b^{2}$的值.

答案： 解: $\because a = \frac{1}{\sqrt{2} - 1} = \sqrt{2} + 1$, $b = \frac{1}{\sqrt{2} + 1} = \sqrt{2} - 1$,
$\therefore a + b = 2\sqrt{2}$, $ab = 2 - 1 = 1$,
$\therefore a^2 - 3ab + b^2 = (a + b)^2 - 5ab = (2\sqrt{2})^2 - 5 × 1 = 3$.

14. (石室联中)已知$a= \frac {1}{2+\sqrt {3}}$,$b= \frac {1}{2-\sqrt {3}}$,试求$a^{2}+b^{2}和a^{2}+3ab+b^{2}$的值.

答案： 解: $a = \frac{1}{2 + \sqrt{3}} = \frac{2 - \sqrt{3}}{(2 + \sqrt{3})(2 - \sqrt{3})} = 2 - \sqrt{3}$,
$b = \frac{1}{2 - \sqrt{3}} = \frac{2 + \sqrt{3}}{(2 - \sqrt{3})(2 + \sqrt{3})} = 2 + \sqrt{3}$,
$a^2 + b^2 = (2 - \sqrt{3})^2 + (2 + \sqrt{3})^2 = (4 - 4\sqrt{3} + 3) + (4 + 4\sqrt{3} + 3) = 7 - 4\sqrt{3} + 7 + 4\sqrt{3} = 14$,
$a + b = (2 - \sqrt{3}) + (2 + \sqrt{3}) = 4$,
$ab = (2 - \sqrt{3})(2 + \sqrt{3}) = 4 - 3 = 1$,
$a^2 + 3ab + b^2 = (a + b)^2 + ab = 4^2 + 1 = 16 + 1 = 17$.

15. (实外西区)已知$x= \frac {1}{2+\sqrt {3}}$,$y= \frac {1}{2-\sqrt {3}}$.
(1)求$x^{2}+y^{2}-xy$的值；
(2)若$x的小数部分为a$,$y的小数部分为b$,求$(a+b)^{2}+\sqrt {(a-b)^{2}}$的值.

答案： 解:
(1) $x = 2 - \sqrt{3}$, $y = 2 + \sqrt{3}$, 原式 $= (x - y)^2 + xy = (2 - \sqrt{3} - 2 - \sqrt{3})^2 + (2 - \sqrt{3})(2 + \sqrt{3}) = 12 + 1 = 13$.
(2) $a = 2 - \sqrt{3}$, $b = \sqrt{3} - 1$, 原式 $= (2 - \sqrt{3} + \sqrt{3} - 1)^2 + \sqrt{(2 - \sqrt{3} - \sqrt{3} + 1)^2} = 1 + \sqrt{(3 - 2\sqrt{3})^2} = 1 + 2\sqrt{3} - 3 = 2\sqrt{3} - 2$.