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2025年寒假零距离七年级数学人教版
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年寒假零距离七年级数学人教版 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。
21. 有这样一道题：“计算$(2x^{4}-4x^{3}y - 2x^{2}y^{2})-(x^{4}-2x^{2}y^{2}+y^{3})+(-x^{4}+4x^{3}y - y^{3})$的值，其中$x = \frac{1}{4}$，$y = - 1$.”甲同学把“$x = \frac{1}{4}$”错抄成“$x = -\frac{1}{4}$”，但他计算的结果也是正确的，你能说明这是为什么吗?
答案：
解：$(2x^{4}-4x^{3}y - 2x^{2}y^{2})-(x^{4}-2x^{2}y^{2}+y^{3})+(-x^{4}+4x^{3}y - y^{3})=2x^{4}-4x^{3}y - 2x^{2}y^{2}-x^{4}+2x^{2}y^{2}-y^{3}-x^{4}+4x^{3}y - y^{3}=-2y^{3}$.
由化简后的代数式可以看出，没有含字母$x$的项存在，所以代数式的值与$x$的取值无关，因此，尽管甲同学把“$x=\frac{1}{4}$”错抄成“$x = -\frac{1}{4}$”，但结果仍是正确的.
由化简后的代数式可以看出，没有含字母$x$的项存在，所以代数式的值与$x$的取值无关，因此，尽管甲同学把“$x=\frac{1}{4}$”错抄成“$x = -\frac{1}{4}$”，但结果仍是正确的.
22. 问题：已知$A = 16a^{2}+a + 15$，$B = 4a^{2}+\frac{1}{2}a+7$，$C = a^{2}+\frac{1}{3}a + 4$. 请你按照上述文字提供的信息：
(1)试比较A与2B的大小；
(2)试比较2B与3C的大小.
(1)试比较A与2B的大小；
(2)试比较2B与3C的大小.
答案：
解：
(1)$A - 2B = 16a^{2}+a + 15-2(4a^{2}+\frac{1}{2}a + 7)=16a^{2}+a + 15 - 8a^{2}-a - 14 = 8a^{2}+1$. 因为$8a^{2}+1＞0$，所以$A＞2B$；
(2)$2B - 3C = 2(4a^{2}+\frac{1}{2}a + 7)-3(a^{2}+\frac{1}{3}a + 4)=8a^{2}+a + 14 - 3a^{2}-a - 12 = 5a^{2}+2＞0$，所以$2B＞3C$.
(1)$A - 2B = 16a^{2}+a + 15-2(4a^{2}+\frac{1}{2}a + 7)=16a^{2}+a + 15 - 8a^{2}-a - 14 = 8a^{2}+1$. 因为$8a^{2}+1＞0$，所以$A＞2B$；
(2)$2B - 3C = 2(4a^{2}+\frac{1}{2}a + 7)-3(a^{2}+\frac{1}{3}a + 4)=8a^{2}+a + 14 - 3a^{2}-a - 12 = 5a^{2}+2＞0$，所以$2B＞3C$.
23. 先阅读提供的材料，再解决相关问题.
$(1+\frac{1}{2})\times(1-\frac{1}{3})=\frac{3}{2}\times\frac{2}{3}=1$；
$(1+\frac{1}{2})\times(1+\frac{1}{4})\times(1-\frac{1}{3})\times(1-\frac{1}{5})=\frac{3}{2}\times\frac{5}{4}\times\frac{2}{3}\times\frac{4}{5}=(\frac{3}{2}\times\frac{2}{3})\times(\frac{5}{4}\times\frac{4}{5})=1\times1 = 1$.
(1)请你求出$(1+\frac{1}{2})\times(1+\frac{1}{4})\times(1+\frac{1}{6})\times(1-\frac{1}{3})\times(1-\frac{1}{5})\times(1-\frac{1}{7})$的结果；
(2)请你运用上述规律求下面式子的值：
$(1+\frac{1}{2})\times(1+\frac{1}{4})\times(1+\frac{1}{6})\times\cdots\times(1+\frac{1}{20})\times(1-\frac{1}{3})\times(1-\frac{1}{5})\times(1-\frac{1}{7})\times\cdots\times(1-\frac{1}{21})$.
$(1+\frac{1}{2})\times(1-\frac{1}{3})=\frac{3}{2}\times\frac{2}{3}=1$；
$(1+\frac{1}{2})\times(1+\frac{1}{4})\times(1-\frac{1}{3})\times(1-\frac{1}{5})=\frac{3}{2}\times\frac{5}{4}\times\frac{2}{3}\times\frac{4}{5}=(\frac{3}{2}\times\frac{2}{3})\times(\frac{5}{4}\times\frac{4}{5})=1\times1 = 1$.
(1)请你求出$(1+\frac{1}{2})\times(1+\frac{1}{4})\times(1+\frac{1}{6})\times(1-\frac{1}{3})\times(1-\frac{1}{5})\times(1-\frac{1}{7})$的结果；
(2)请你运用上述规律求下面式子的值：
$(1+\frac{1}{2})\times(1+\frac{1}{4})\times(1+\frac{1}{6})\times\cdots\times(1+\frac{1}{20})\times(1-\frac{1}{3})\times(1-\frac{1}{5})\times(1-\frac{1}{7})\times\cdots\times(1-\frac{1}{21})$.
答案：
解：
(1)原式$=\frac{3}{2}\times\frac{5}{4}\times\frac{7}{6}\times\frac{2}{3}\times\frac{4}{5}\times\frac{6}{7}=(\frac{3}{2}\times\frac{2}{3})\times(\frac{5}{4}\times\frac{4}{5})\times(\frac{7}{6}\times\frac{6}{7}) = 1\times1\times1 = 1$；
(2)原式$=\frac{3}{2}\times\frac{5}{4}\times\frac{7}{6}\times\cdots\times\frac{21}{20}\times\frac{2}{3}\times\frac{4}{5}\times\frac{6}{7}\times\cdots\times\frac{20}{21}=(\frac{3}{2}\times\frac{2}{3})\times(\frac{5}{4}\times\frac{4}{5})\times(\frac{7}{6}\times\frac{6}{7})\times\cdots\times(\frac{21}{20}\times\frac{20}{21}) = 1\times1\times1\times\cdots\times1$
(1)原式$=\frac{3}{2}\times\frac{5}{4}\times\frac{7}{6}\times\frac{2}{3}\times\frac{4}{5}\times\frac{6}{7}=(\frac{3}{2}\times\frac{2}{3})\times(\frac{5}{4}\times\frac{4}{5})\times(\frac{7}{6}\times\frac{6}{7}) = 1\times1\times1 = 1$；
(2)原式$=\frac{3}{2}\times\frac{5}{4}\times\frac{7}{6}\times\cdots\times\frac{21}{20}\times\frac{2}{3}\times\frac{4}{5}\times\frac{6}{7}\times\cdots\times\frac{20}{21}=(\frac{3}{2}\times\frac{2}{3})\times(\frac{5}{4}\times\frac{4}{5})\times(\frac{7}{6}\times\frac{6}{7})\times\cdots\times(\frac{21}{20}\times\frac{20}{21}) = 1\times1\times1\times\cdots\times1$
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