《2025年学霸题中题八年级数学上册人教版》

13. 阅读理解题.
计算:$(-0.125)^{4}×8^{5}$.
解:原式$=(-\frac {1}{8})^{4}×8^{5}=(\frac {1}{8})^{4}×8^{4}×8$
$=(\frac {1}{8}×8)^{4}×8=8$.
请根据上面的解题思路计算下列各题:
(1)$(-3)^{1001}×(-\frac {1}{3})^{999}$;
解:原式$=(-3)^{2}×(-3)^{999}×(-\frac {1}{3})^{999}=9×[(-3)×(-\frac {1}{3})]^{999}=9×1^{999}=$9.
(2)$0.125^{3}×0.25^{3}×2^{6}×2^{12}$;
解:原式$=0.125^{3}×0.25^{3}×4^{3}×16^{3}=(0.125×0.25×4×16)^{3}=2^{3}=$8.
(3)化简$(\frac {b}{a})^{n}(-\frac {a}{b})^{n+2}$.
解:原式$=(\frac {b}{a})^{n}(-\frac {a}{b})^{n}(-\frac {a}{b})^{2}=[\frac {b}{a}\cdot (-\frac {a}{b})]^{n}\cdot (\frac {a}{b})^{2}=$$(-1)^{n}(\frac {a}{b})^{2}$.

答案：
(1)原式$=(-3)^{2}×(-3)^{999}×(-\frac {1}{3})^{999}=9×[(-3)×(-\frac {1}{3})]^{999}=9×1^{999}=9.$
(2)原式$=0.125^{3}×0.25^{3}×4^{3}×16^{3}=(0.125×0.25×4×16)^{3}=2^{3}=8.$
(3)原式$=(\frac {b}{a})^{n}(-\frac {a}{b})^{n}(-\frac {a}{b})^{2}=[\frac {b}{a}\cdot (-\frac {a}{b})]^{n}\cdot (\frac {a}{b})^{2}=(-1)^{n}(\frac {a}{b})^{2}.$

14. 改编题 若$25^{x}=2025,81^{y}=2025$,试说明代数式xy与$x+y$之间的关系.
由条件可得$(25^{x})^{y}=2025^{y},(81^{y})^{x}=2025^{x}$，可得$(25^{x})^{y}\cdot (81^{y})^{x}=25^{xy}\cdot 81^{xy}=2025^{y}\cdot 2025^{x}=2025^{x+y}$，而$2025=25×81,\therefore 25^{xy}\cdot 81^{xy}=2025^{xy}=2025^{x+y},\therefore xy=x+y.$

答案： 由条件可得$(25^{x})^{y}=2025^{y},(81^{y})^{x}=2025^{x}$，可得$(25^{x})^{y}\cdot (81^{y})^{x}=25^{xy}\cdot 81^{xy}=2025^{y}\cdot 2025^{x}=2025^{x+y}$，而$2025=25×81,\therefore 25^{xy}\cdot 81^{xy}=2025^{xy}=2025^{x+y},\therefore xy=x+y.$

15. 用所学知识完成下列题目:
(1)若$2^{a}=3,2^{b}=6,2^{c}=12$,直接写出a,b,c之间的数量关系:$a+c=2b$;
(2)若$2^{a}=6,4^{b}=12,16^{c}=8$,试确定a,b,c之间的数量关系,并说明理由;$a+4c=2+2b$
(3)若$a^{5}=2,b^{5}=3,c^{5}=72$,试确定a,b,c之间的数量关系,并说明理由.$c=a^{3}b^{2}$

答案：
(1)$a+c=2b$ 解析：$\because 2^{a}\cdot 2^{c}=2^{a+c}=3×12=36,2^{b}\cdot 2^{b}=2^{2b}=6×6=36,\therefore 2^{a+c}=2^{2b}$，即$a+c=2b.$
(2)$a+4c=2+2b$，理由：$\because 6×8=4×12,\therefore 2^{a}×16^{c}=4×4^{b}.$又$\because 4^{b}=2^{2b}=12,16^{c}=2^{4c}=8,\therefore a+4c=2+2b.$
(3)$c=a^{3}b^{2}$，理由：$\because c^{5}=72=2^{3}×3^{2}=(a^{5})^{3}\cdot (b^{5})^{2}=(a^{3}b^{2})^{5},\therefore c=a^{3}b^{2}.$

16. 求$3^{1001}×9^{1002}×27^{1003}$的末位数字.

答案： $3^{1}=3,3^{2}=9,3^{3}=27,3^{4}=81,3^{5}=243,3^{6}=729,3^{7}=2187,3^{8}=6561,...$，观察上式可得$3^{n}$(n为正整数)的末位数字每4个一循环，$3^{1001}×9^{1002}×27^{1003}=3^{1001}×(3^{2})^{1002}×(3^{3})^{1003}=3^{1001+2004+3009}=3^{6014},6014÷4=1503... 2$，
∴其末位数字和$3^{2}$的末位数字一样，即为9.

17. $5^{2}\cdot 3^{2n+1}\cdot 2^{n}-3^{n}\cdot 6^{n+2}$(n为正整数)能被13整除吗?

答案： $5^{2}\cdot 3^{2n+1}\cdot 2^{n}-3^{n}\cdot 6^{n+2}=5^{2}\cdot (3^{2n}\cdot 3)\cdot 2^{n}-3^{n}\cdot (6^{n}\cdot 6^{2})=75\cdot 3^{2n}\cdot 2^{n}-36\cdot 3^{n}\cdot 6^{n}=75\cdot 18^{n}-36\cdot 18^{n}=39\cdot 18^{n}=13×3\cdot 18^{n}.$$\because 3\cdot 18^{n}$是整数，$\therefore 5^{2}\cdot 3^{2n+1}\cdot 2^{n}-3^{n}\cdot 6^{n+2}$能被13整除.