《2025年综合应用创新题典中点八年级数学上册北师大版》

14.(8分)(1)若$|2x-4|+(y+3)^{2}+\sqrt {x+y+z}= 0$,求$x-2y+z$的平方根;$\pm 3$
(2)先化简,再求值:$(\sqrt {2x}+\sqrt {y})(\sqrt {2x}-\sqrt {y})-(\sqrt {2x}-\sqrt {y})^{2}$,其中$x= \frac {3}{4},y= \frac {1}{2}$.$\sqrt{3}-1$

答案： [解]
(1)因为$\vert 2x - 4\vert + (y + 3)^{2} + \sqrt{x + y + z} = 0$，所以$2x - 4 = 0$，$y + 3 = 0$，$x + y + z = 0$，所以$x = 2$，$y = -3$，$z = 1$，所以$x - 2y + z = 2 + 6 + 1 = 9$，所以$x - 2y + z$的平方根为$\pm 3$。
(2)原式$ = (\sqrt{2x})^{2} - (\sqrt{y})^{2} - (\sqrt{2x} - \sqrt{y})^{2} = 2x - y - 2x + 2\sqrt{2xy} - y = 2\sqrt{2xy} - 2y$。当$x = \frac{3}{4}$，$y = \frac{1}{2}$时，原式$ = 2 \times \sqrt{2 \times \frac{3}{4} \times \frac{1}{2}} - 2 \times \frac{1}{2} = \sqrt{3} - 1$。

15.(10分)已知$x= \frac {1}{3+2\sqrt {2}},y= \frac {1}{3-2\sqrt {2}}$.
(1)求$x^{2}+y^{2}+xy$的值;35
(2)若x的小数部分是m,y的小数部分是n,求$(m+n)^{2026}-\sqrt [3]{(m-n)^{3}}$的值.$4\sqrt{2}-4$

答案： [解]
(1)因为$x = \frac{1}{3 + 2\sqrt{2}} = \frac{3 - 2\sqrt{2}}{(3 + 2\sqrt{2})(3 - 2\sqrt{2})} = 3 - 2\sqrt{2}$，$y = \frac{1}{3 - 2\sqrt{2}} = \frac{3 + 2\sqrt{2}}{(3 + 2\sqrt{2})(3 - 2\sqrt{2})} = 3 + 2\sqrt{2}$，所以$x^{2} + y^{2} + xy = x^{2} + y^{2} + 2xy - xy = (x + y)^{2} - xy = (3 - 2\sqrt{2} + 3 + 2\sqrt{2})^{2} - (3 + 2\sqrt{2})(3 - 2\sqrt{2}) = 6^{2} - (9 - 8) = 36 - 1 = 35$。
(2)因为$4 ＜ 8 ＜ 9$，所以$\sqrt{4} ＜ \sqrt{8} ＜ \sqrt{9}$，即$2 ＜ 2\sqrt{2} ＜ 3$，所以$0 ＜ 3 - 2\sqrt{2} ＜ 1$，$5 ＜ 3 + 2\sqrt{2} ＜ 6$。由
(1)可知，$x = 3 - 2\sqrt{2}$，$y = 3 + 2\sqrt{2}$，所以$0 ＜ x ＜ 1$，$5 ＜ y ＜ 6$。因为$x$的小数部分是$m$，$y$的小数部分是$n$，所以$m = 3 - 2\sqrt{2}$，$n = 3 + 2\sqrt{2} - 5 = 2\sqrt{2} - 2$。所以$(m + n)^{2026} - \sqrt[3]{(m - n)^{3}} = (3 - 2\sqrt{2} + 2\sqrt{2} - 2)^{2026} - \sqrt[3]{(3 - 2\sqrt{2} - 2\sqrt{2} + 2)^{3}} = 1^{2026} - \sqrt[3]{(5 - 4\sqrt{2})^{3}} = 1 - (5 - 4\sqrt{2}) = 1 - 5 + 4\sqrt{2} = 4\sqrt{2} - 4$。

16.(12分)情境题·生活应用 现有两块同样大小的长方形木板①,②,甲木工采用如图①所示的方式,在长方形木板①上截出三个面积分别为$4dm^{2},8dm^{2}和18dm^{2}$的正方形木板A,B,C.
(1)木板①中截出的正方形木板A的边长为2dm,B的边长为$2\sqrt{2}$dm,C的边长为$3\sqrt{2}$dm;
(2)求木板①中剩余部分(阴影部分)的面积;
(3)乙木工想采用如图②所示的方式,在长方形木板②上截出两个面积均为$16dm^{2}$的正方形木板,请你判断能否截出,并说明理由.

答案： [解]
(1)$2$；$2\sqrt{2}$；$3\sqrt{2}$
(2)因为正方形木板$A$的边长为$2dm$，正方形木板$B$的边长为$2\sqrt{2}dm$，正方形木板$C$的边长为$3\sqrt{2}dm$，所以长方形木板①的长为$5\sqrt{2}dm$，宽为$(2 + 2\sqrt{2})dm$，所以剩余部分(阴影部分)的面积为$5\sqrt{2}(2 + 2\sqrt{2}) - 4 - 8 - 18 = (10\sqrt{2} - 10)(dm^{2})$。
(3)不能截出。
理由：因为$\sqrt{16} = 4(dm)$，$2 \times 4 = 8(dm)$，所以两个正方形木板放在一起的宽为$4dm$，长为$8dm$。由
(2)可得长方形木板的长为$5\sqrt{2}dm$，宽为$(2 + 2\sqrt{2})dm$。因为$2 + 2\sqrt{2} ＞ 4$，但$5\sqrt{2} ＜ 8$，所以不能截出。

17.(12分)新视角 规律探究题 观察下列一组等式,然后解答问题:$(\sqrt {2}+1)(\sqrt {2}-1)= 1,(\sqrt {3}+\sqrt {2})(\sqrt {3}-\sqrt {2})= 1,(\sqrt {4}+\sqrt {3})(\sqrt {4}-\sqrt {3})= 1,(\sqrt {5}+\sqrt {4})(\sqrt {5}-\sqrt {4})= 1,... $.
(1)利用上面的规律,计算:$\frac {1}{\sqrt {2}+1}+\frac {1}{\sqrt {3}+\sqrt {2}}+\frac {1}{\sqrt {4}+\sqrt {3}}+... +\frac {1}{\sqrt {2025}+\sqrt {2024}}$=44;
(2)请利用上面的规律,比较$\sqrt {99}-\sqrt {98}与\sqrt {98}-\sqrt {97}$的大小.$\sqrt{99} - \sqrt{98} ＜ \sqrt{98} - \sqrt{97}$

答案： [解]
(1)原式$ = \frac{\sqrt{2} - 1}{(\sqrt{2} + 1)(\sqrt{2} - 1)} + \frac{\sqrt{3} - \sqrt{2}}{(\sqrt{3} + \sqrt{2})(\sqrt{3} - \sqrt{2})} + \frac{\sqrt{4} - \sqrt{3}}{(\sqrt{4} + \sqrt{3})(\sqrt{4} - \sqrt{3})} + \cdots + \frac{\sqrt{2025} - \sqrt{2024}}{(\sqrt{2025} + \sqrt{2024})(\sqrt{2025} - \sqrt{2024})} = \sqrt{2} - 1 + \sqrt{3} - \sqrt{2} + \sqrt{4} - \sqrt{3} + \cdots + \sqrt{2025} - \sqrt{2024} = -1 + \sqrt{2025} = 44$。
(2)$\sqrt{99} - \sqrt{98} = \frac{(\sqrt{99} + \sqrt{98})(\sqrt{99} - \sqrt{98})}{\sqrt{99} + \sqrt{98}} = \frac{1}{\sqrt{99} + \sqrt{98}}$，$\sqrt{98} - \sqrt{97} = \frac{(\sqrt{98} + \sqrt{97})(\sqrt{98} - \sqrt{97})}{\sqrt{98} + \sqrt{97}} = \frac{1}{\sqrt{98} + \sqrt{97}}$。因为$\frac{1}{\sqrt{99} + \sqrt{98}} ＜ \frac{1}{\sqrt{98} + \sqrt{97}}$，所以$\sqrt{99} - \sqrt{98} ＜ \sqrt{98} - \sqrt{97}$。