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2025年暑假作业内蒙古大学出版社八年级数学
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年暑假作业内蒙古大学出版社八年级数学 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。
请你参考上述算法,运用平方差公式简便计算:
(1)$\frac {19}{2}×\frac {21}{2};$
(2)$(2021\sqrt {3}+2021\sqrt {2})(\sqrt {3}-\sqrt {2}).$
(1)$\frac {19}{2}×\frac {21}{2};$
(2)$(2021\sqrt {3}+2021\sqrt {2})(\sqrt {3}-\sqrt {2}).$
答案：
1.解:
(1)原式=$\frac{1}{4}$×(20−1)×(20+1)
=$\frac{1}{4}$×(20²−1²)=$\frac{1}{4}$×(400−1)=$\frac{399}{4}$;
(2)原式=2021×($\sqrt{3}$+$\sqrt{2}$)×($\sqrt{3}$−$\sqrt{2}$)
=2021×(3−2)=2021.
(1)原式=$\frac{1}{4}$×(20−1)×(20+1)
=$\frac{1}{4}$×(20²−1²)=$\frac{1}{4}$×(400−1)=$\frac{399}{4}$;
(2)原式=2021×($\sqrt{3}$+$\sqrt{2}$)×($\sqrt{3}$−$\sqrt{2}$)
=2021×(3−2)=2021.
2. 阅读下列材料,然后解答问题:
在进行二次根式的化简与运算时,我们有时会碰上如:$\frac {3}{\sqrt {5}},\sqrt {\frac {2}{3}},\frac {2}{\sqrt {3}+1}$一样的式子.其实我们还可以将其进一步化简:
$\frac {3}{\sqrt {5}}= \frac {3×\sqrt {5}}{\sqrt {5}×\sqrt {5}}= \frac {3\sqrt {5}}{5}$. (一)
$\sqrt {\frac {2}{3}}= \frac {\sqrt {2×3}}{\sqrt {3×3}}= \frac {\sqrt {6}}{3}$. (二)
$\frac {2}{\sqrt {3}+1}= \frac {2(\sqrt {3}-1)}{(\sqrt {3}+1)(\sqrt {3}-1)}= \frac {2(\sqrt {3}-1)}{(\sqrt {3})^{2}-1}= \sqrt {3}-1$. (三)
以上这种化简的步骤叫做分母有理化.
$\frac {2}{\sqrt {3}+1}$还可以用以下方法化简:
$\frac {2}{\sqrt {3}+1}= \frac {3-1}{\sqrt {3}+1}= \frac {(\sqrt {3})^{2}-1}{\sqrt {3}+1}= \frac {(\sqrt {3}+1)(\sqrt {3}-1)}{\sqrt {3}+1}= \sqrt {3}-1$. (四)
请解答下列问题:
(1)请用不同的方法化简$\frac {2}{\sqrt {5}+\sqrt {3}}:$
①参照(三)式得$\frac {2}{\sqrt {5}+\sqrt {3}}= $____;
②参照(四)式得$\frac {2}{\sqrt {5}+\sqrt {3}}= $____= ____=____.
(2)化简:$\frac {2}{\sqrt {3}+1}+\frac {2}{\sqrt {5}+\sqrt {3}}+\frac {2}{\sqrt {7}+\sqrt {5}}$. (保留过程)
(3)猜想:$\frac {1}{\sqrt {3}+1}+\frac {1}{\sqrt {5}+\sqrt {3}}+\frac {1}{\sqrt {7}+\sqrt {5}}+... +\frac {1}{\sqrt {2n+1}+\sqrt {2n-1}}$的值. (直接写出结论)
在进行二次根式的化简与运算时,我们有时会碰上如:$\frac {3}{\sqrt {5}},\sqrt {\frac {2}{3}},\frac {2}{\sqrt {3}+1}$一样的式子.其实我们还可以将其进一步化简:
$\frac {3}{\sqrt {5}}= \frac {3×\sqrt {5}}{\sqrt {5}×\sqrt {5}}= \frac {3\sqrt {5}}{5}$. (一)
$\sqrt {\frac {2}{3}}= \frac {\sqrt {2×3}}{\sqrt {3×3}}= \frac {\sqrt {6}}{3}$. (二)
$\frac {2}{\sqrt {3}+1}= \frac {2(\sqrt {3}-1)}{(\sqrt {3}+1)(\sqrt {3}-1)}= \frac {2(\sqrt {3}-1)}{(\sqrt {3})^{2}-1}= \sqrt {3}-1$. (三)
以上这种化简的步骤叫做分母有理化.
$\frac {2}{\sqrt {3}+1}$还可以用以下方法化简:
$\frac {2}{\sqrt {3}+1}= \frac {3-1}{\sqrt {3}+1}= \frac {(\sqrt {3})^{2}-1}{\sqrt {3}+1}= \frac {(\sqrt {3}+1)(\sqrt {3}-1)}{\sqrt {3}+1}= \sqrt {3}-1$. (四)
请解答下列问题:
(1)请用不同的方法化简$\frac {2}{\sqrt {5}+\sqrt {3}}:$
①参照(三)式得$\frac {2}{\sqrt {5}+\sqrt {3}}= $____;
②参照(四)式得$\frac {2}{\sqrt {5}+\sqrt {3}}= $____= ____=____.
(2)化简:$\frac {2}{\sqrt {3}+1}+\frac {2}{\sqrt {5}+\sqrt {3}}+\frac {2}{\sqrt {7}+\sqrt {5}}$. (保留过程)
(3)猜想:$\frac {1}{\sqrt {3}+1}+\frac {1}{\sqrt {5}+\sqrt {3}}+\frac {1}{\sqrt {7}+\sqrt {5}}+... +\frac {1}{\sqrt {2n+1}+\sqrt {2n-1}}$的值. (直接写出结论)
答案：
2.解:
(1)①参照(三)式得$\frac{2}{\sqrt{5}+\sqrt{3}}$ =$\frac{2(\sqrt{5}-\sqrt{3})}{(\sqrt{5}+\sqrt{3})(\sqrt{5}-\sqrt{3})}$ =$\frac{2(\sqrt{5}-\sqrt{3})}{(\sqrt{5})^{2}-(\sqrt{3})^{2}}$ =$\sqrt{5}$−$\sqrt{3}$; ②参照(四)式得$\frac{2}{\sqrt{5}+\sqrt{3}}=\frac{(\sqrt{5})^{2}-(\sqrt{3})^{2}}{\sqrt{5}+\sqrt{3}}=\frac{(\sqrt{5}+\sqrt{3})(\sqrt{5}-\sqrt{3})}{\sqrt{5}+\sqrt{3}}=\sqrt{5}-\sqrt{3}$;
(2)化简:$\frac{2}{\sqrt{3}+1}+\frac{2}{\sqrt{5}+\sqrt{3}}+\frac{2}{\sqrt{7}+\sqrt{5}}$ =$\sqrt{3}$−1+$\sqrt{5}$−$\sqrt{3}$+$\sqrt{7}$−$\sqrt{5}$=$\sqrt{7}$−1;
(3)猜想:$\frac{1}{\sqrt{3}+1}+\frac{1}{\sqrt{5}+\sqrt{3}}+\frac{1}{\sqrt{7}+\sqrt{5}}+\cdots+\frac{1}{\sqrt{2n+1}+\sqrt{2n−1}}=\frac{1}{2}(\sqrt{2n+1}-1)$.
(1)①参照(三)式得$\frac{2}{\sqrt{5}+\sqrt{3}}$ =$\frac{2(\sqrt{5}-\sqrt{3})}{(\sqrt{5}+\sqrt{3})(\sqrt{5}-\sqrt{3})}$ =$\frac{2(\sqrt{5}-\sqrt{3})}{(\sqrt{5})^{2}-(\sqrt{3})^{2}}$ =$\sqrt{5}$−$\sqrt{3}$; ②参照(四)式得$\frac{2}{\sqrt{5}+\sqrt{3}}=\frac{(\sqrt{5})^{2}-(\sqrt{3})^{2}}{\sqrt{5}+\sqrt{3}}=\frac{(\sqrt{5}+\sqrt{3})(\sqrt{5}-\sqrt{3})}{\sqrt{5}+\sqrt{3}}=\sqrt{5}-\sqrt{3}$;
(2)化简:$\frac{2}{\sqrt{3}+1}+\frac{2}{\sqrt{5}+\sqrt{3}}+\frac{2}{\sqrt{7}+\sqrt{5}}$ =$\sqrt{3}$−1+$\sqrt{5}$−$\sqrt{3}$+$\sqrt{7}$−$\sqrt{5}$=$\sqrt{7}$−1;
(3)猜想:$\frac{1}{\sqrt{3}+1}+\frac{1}{\sqrt{5}+\sqrt{3}}+\frac{1}{\sqrt{7}+\sqrt{5}}+\cdots+\frac{1}{\sqrt{2n+1}+\sqrt{2n−1}}=\frac{1}{2}(\sqrt{2n+1}-1)$.
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