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2025年文轩图书假期生活指导暑八年级数学通用版
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年文轩图书假期生活指导暑八年级数学通用版 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。
1. 利用平方差公式可以进行简便计算:
例1:$99×101= (100-1)(100+1)= 100^{2}-1^{2}= 10000-1= 9999;$
例2:$39×410= 39×41×10= (40-1)(40+1)×10= (40^{2}-1^{2})×10= (1600-1)×10= 1599×10= 15990.$
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)$(2021\sqrt {3}+2021\sqrt {2})(\sqrt {3}-\sqrt {2})=$
例1:$99×101= (100-1)(100+1)= 100^{2}-1^{2}= 10000-1= 9999;$
例2:$39×410= 39×41×10= (40-1)(40+1)×10= (40^{2}-1^{2})×10= (1600-1)×10= 1599×10= 15990.$
请你参考上述算法,运用平方差公式简便计算:
(1)$\frac {19}{2}×\frac {21}{2}=$
$\frac{1}{4}$×(20−1)×(20+1)=$\frac{1}{4}$×(20²−1²)=$\frac{1}{4}$×(400−1)=$\frac{399}{4}$
;(2)$(2021\sqrt {3}+2021\sqrt {2})(\sqrt {3}-\sqrt {2})=$
2021×($\sqrt{3}$+$\sqrt{2}$)×($\sqrt{3}$−$\sqrt{2}$)=2021×(3−2)=2021
.
答案：
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$\frac {2}{\sqrt {5}+\sqrt {3}}=$
$\frac{2(\sqrt{5}-\sqrt{3})}{(\sqrt{5}+\sqrt{3})(\sqrt{5}-\sqrt{3})}=\frac{2(\sqrt{5}-\sqrt{3})}{(\sqrt{5})²−(\sqrt{3})²}=\sqrt{5}-\sqrt{3}$
;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:$\frac {2}{\sqrt {3}+1}+\frac {2}{\sqrt {5}+\sqrt {3}}+\frac {2}{\sqrt {7}+\sqrt {5}}$. (保留过程)
$\frac{2}{\sqrt{3}+1}+\frac{2}{\sqrt{5}+\sqrt{3}}+\frac{2}{\sqrt{7}+\sqrt{5}}=\sqrt{3}-1+\sqrt{5}-\sqrt{3}+\sqrt{7}-\sqrt{5}=\sqrt{7}-1$
(3)猜想:$\frac {1}{\sqrt {3}+1}+\frac {1}{\sqrt {5}+\sqrt {3}}+\frac {1}{\sqrt {7}+\sqrt {5}}+... +\frac {1}{\sqrt {2n+1}+\sqrt {2n-1}}$的值. (直接写出结论)
$\frac{1}{2}(\sqrt{2n+1}-1)$
答案：
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:$\frac{1}{\sqrt{3}+1}$+$\frac{1}{\sqrt{5}+\sqrt{3}}$+$\frac{1}{\sqrt{7}+\sqrt{5}}$+…+$\frac{1}{\sqrt{2n+1}+\sqrt{2n−1}}$=$\frac{1}{2}$($\sqrt{2n+1}$−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:$\frac{1}{\sqrt{3}+1}$+$\frac{1}{\sqrt{5}+\sqrt{3}}$+$\frac{1}{\sqrt{7}+\sqrt{5}}$+…+$\frac{1}{\sqrt{2n+1}+\sqrt{2n−1}}$=$\frac{1}{2}$($\sqrt{2n+1}$−1).
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