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2025年鲁人泰斗假期好时光八年级数学青岛版
注：目前有些书本章节名称可能整理的还不是很完善，但都是按照顺序排列的，请同学们按照顺序仔细查找。练习册 2025年鲁人泰斗假期好时光八年级数学青岛版 答案主要是用来给同学们做完题方便对答案用的，请勿直接抄袭。
17. (10分)已知$x= \frac {2}{3+\sqrt {5}},y= \frac {2}{3-\sqrt {5}}$,求$2x^{2}-xy+2y^{2}$的值.
解：$\because x=\frac{2}{3+\sqrt{5}}=\frac{3-\sqrt{5}}{2}$，$y=\frac{2}{3-\sqrt{5}}=\frac{3+\sqrt{5}}{2}$，
$\therefore x+y=$
$\therefore 2x^{2}-xy+2y^{2}=2(x+y)^{2}-xy-4xy=2×3^{2}-5×1=$
解：$\because x=\frac{2}{3+\sqrt{5}}=\frac{3-\sqrt{5}}{2}$，$y=\frac{2}{3-\sqrt{5}}=\frac{3+\sqrt{5}}{2}$，
$\therefore x+y=$
3
，$xy=$1
.$\therefore 2x^{2}-xy+2y^{2}=2(x+y)^{2}-xy-4xy=2×3^{2}-5×1=$
13
.
答案：
解：$\because x=\frac{2}{3+\sqrt{5}}=\frac{3-\sqrt{5}}{2}$，$y=\frac{2}{3-\sqrt{5}}=\frac{3+\sqrt{5}}{2}$，
$\therefore x+y=3$，$xy=1$.
$\therefore 2x^{2}-xy+2y^{2}=2(x+y)^{2}-xy-4xy=2×3^{2}-5×1=13$.
$\therefore x+y=3$，$xy=1$.
$\therefore 2x^{2}-xy+2y^{2}=2(x+y)^{2}-xy-4xy=2×3^{2}-5×1=13$.
18. (10分)有一块矩形木块,木工采用如图方式,在木板上截出两个面积分别为$18dm^{2}和32dm^{2}$的正方形木板,求剩余木料的面积.
解：$\because$ 两个正方形木板的面积分别为$18dm^{2}$和$32dm^{2}$，
$\therefore$ 这两个正方形的边长分别为
$\sqrt{18}=$
$\therefore$ 剩余木料的面积为$(4\sqrt{2}-3\sqrt{2})×3\sqrt{2}=$
解：$\because$ 两个正方形木板的面积分别为$18dm^{2}$和$32dm^{2}$，
$\therefore$ 这两个正方形的边长分别为
$\sqrt{18}=$
$3\sqrt{2}$
$(dm)$，$\sqrt{32}=$$4\sqrt{2}$
$(dm)$.$\therefore$ 剩余木料的面积为$(4\sqrt{2}-3\sqrt{2})×3\sqrt{2}=$
$\sqrt{2}×3\sqrt{2}=6$
$(dm^{2})$.
答案：
解：$\because$ 两个正方形木板的面积分别为$18dm^{2}$和$32dm^{2}$，
$\therefore$ 这两个正方形的边长分别为
$\sqrt{18}=3\sqrt{2}(dm)$，$\sqrt{32}=4\sqrt{2}(dm)$.
$\therefore$ 剩余木料的面积为$(4\sqrt{2}-3\sqrt{2})×3\sqrt{2}=$
$\sqrt{2}×3\sqrt{2}=6(dm^{2})$.
$\therefore$ 这两个正方形的边长分别为
$\sqrt{18}=3\sqrt{2}(dm)$，$\sqrt{32}=4\sqrt{2}(dm)$.
$\therefore$ 剩余木料的面积为$(4\sqrt{2}-3\sqrt{2})×3\sqrt{2}=$
$\sqrt{2}×3\sqrt{2}=6(dm^{2})$.
19. (10分)佳佳给出了$\sqrt {6}×2\sqrt {3}-\sqrt {24}÷\sqrt {3}$的解题过程：
$\sqrt {6}×2\sqrt {3}-\sqrt {24}÷\sqrt {3}$
$=2\sqrt {6×3}-\sqrt {\frac {24}{3}}$……①
$=2\sqrt {18}-\sqrt {8}$……②
$=(2-1)\sqrt {18-8}$……③
$=\sqrt {10}$……④
(1)佳佳从
(2)请你给出正确的解题过程.
$\sqrt {6}×2\sqrt {3}-\sqrt {24}÷\sqrt {3}$
$=2\sqrt {6×3}-\sqrt {\frac {24}{3}}$……①
$=2\sqrt {18}-\sqrt {8}$……②
$=(2-1)\sqrt {18-8}$……③
$=\sqrt {10}$……④
(1)佳佳从
③
步开始产生错误；(2)请你给出正确的解题过程.
答案：
解：
(1) ③
(2) 原式$=2\sqrt{6×3}-\sqrt{24×\frac{1}{3}}$
$=2\sqrt{18}-\sqrt{8}$
$=6\sqrt{2}-2\sqrt{2}$
$=4\sqrt{2}$.
(1) ③
(2) 原式$=2\sqrt{6×3}-\sqrt{24×\frac{1}{3}}$
$=2\sqrt{18}-\sqrt{8}$
$=6\sqrt{2}-2\sqrt{2}$
$=4\sqrt{2}$.
20. (10分)观察下列等式：
①$\frac {1}{\sqrt {2}+1}= \frac {\sqrt {2}-1}{(\sqrt {2}+1)(\sqrt {2}-1)}= \sqrt {2}-1;$
②$\frac {1}{\sqrt {3}+\sqrt {2}}= \frac {\sqrt {3}-\sqrt {2}}{(\sqrt {3}+\sqrt {2})(\sqrt {3}-\sqrt {2})}= \sqrt {3}-\sqrt {2};$
③$\frac {1}{\sqrt {4}+\sqrt {3}}= \frac {\sqrt {4}-\sqrt {3}}{(\sqrt {4}+\sqrt {3})(\sqrt {4}-\sqrt {3})}= \sqrt {4}-\sqrt {3};$
……
回答下列问题：
(1)利用你观察到的规律：
化简：$\frac {1}{\sqrt {23}+\sqrt {22}}= $
(2)计算：$\frac {1}{1+\sqrt {2}}+\frac {1}{\sqrt {2}+\sqrt {3}}+\frac {1}{\sqrt {3}+2}+... +\frac {1}{\sqrt {99}+\sqrt {100}}.$
$\frac {1}{1+\sqrt {2}}+\frac {1}{\sqrt {2}+\sqrt {3}}+\frac {1}{\sqrt {3}+2}+\cdots+\frac {1}{\sqrt {99}+\sqrt {100}}$
$=\sqrt{2}-1+\sqrt{3}-\sqrt{2}+2-\sqrt{3}+\cdots+\sqrt{100}-\sqrt{99}$
$=\sqrt{100}-1=9$.
①$\frac {1}{\sqrt {2}+1}= \frac {\sqrt {2}-1}{(\sqrt {2}+1)(\sqrt {2}-1)}= \sqrt {2}-1;$
②$\frac {1}{\sqrt {3}+\sqrt {2}}= \frac {\sqrt {3}-\sqrt {2}}{(\sqrt {3}+\sqrt {2})(\sqrt {3}-\sqrt {2})}= \sqrt {3}-\sqrt {2};$
③$\frac {1}{\sqrt {4}+\sqrt {3}}= \frac {\sqrt {4}-\sqrt {3}}{(\sqrt {4}+\sqrt {3})(\sqrt {4}-\sqrt {3})}= \sqrt {4}-\sqrt {3};$
……
回答下列问题：
(1)利用你观察到的规律：
化简：$\frac {1}{\sqrt {23}+\sqrt {22}}= $
$\sqrt{23}-\sqrt{22}$
；(2)计算：$\frac {1}{1+\sqrt {2}}+\frac {1}{\sqrt {2}+\sqrt {3}}+\frac {1}{\sqrt {3}+2}+... +\frac {1}{\sqrt {99}+\sqrt {100}}.$
$\frac {1}{1+\sqrt {2}}+\frac {1}{\sqrt {2}+\sqrt {3}}+\frac {1}{\sqrt {3}+2}+\cdots+\frac {1}{\sqrt {99}+\sqrt {100}}$
$=\sqrt{2}-1+\sqrt{3}-\sqrt{2}+2-\sqrt{3}+\cdots+\sqrt{100}-\sqrt{99}$
$=\sqrt{100}-1=9$.
答案：
解：
(1) $\sqrt{23}-\sqrt{22}$
(2) $\frac{1}{1+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+\frac{1}{\sqrt{3}+2}+\cdots+\frac{1}{\sqrt{99}+\sqrt{100}}$
$=\sqrt{2}-1+\sqrt{3}-\sqrt{2}+2-\sqrt{3}+\cdots+\sqrt{100}-\sqrt{99}$
$=\sqrt{100}-1=9$.
(1) $\sqrt{23}-\sqrt{22}$
(2) $\frac{1}{1+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+\frac{1}{\sqrt{3}+2}+\cdots+\frac{1}{\sqrt{99}+\sqrt{100}}$
$=\sqrt{2}-1+\sqrt{3}-\sqrt{2}+2-\sqrt{3}+\cdots+\sqrt{100}-\sqrt{99}$
$=\sqrt{100}-1=9$.
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