（1）一个暖瓶与一个水杯分别是多少元？
（2）甲、乙两家商场同时出售同样的暖瓶和水杯，为了迎接新年，两家商场都在搞促销活动，甲商场规定：这两种商品都打九折；乙商场规定：买一个暖瓶赠送一个水杯．若某人想要买4个暖瓶和15个水杯，请问选择哪家商场购买更合算，并说明理由． 【分析】（1）设一个暖瓶x元，一个水杯y元，根据“购买一个暖瓶、一个水杯共需100元，购买两个暖瓶、三个水杯共需230元”，即可得出关于x，y的二元一次方程组，解之即可得出结论；
（2）根据两商城的促销方案，分别求出到两商城购买所需费用，比较后即可得出结论． 解：（1）设一个暖瓶x元，一个水杯y元， 根据题意得：解得：
．
， 答：一个暖瓶70元，一个水杯30元
（2）若到甲商场购买，则所需的钱数为：（4×70+15×30）×90%＝657（元）， 若到乙商场购买，则所需的钱数为：4×70+（15﹣4）×30＝610（元）． ∵657＞610，
∴到乙家商场购买更合算．
【点评】本题考查了二元一次方程组的应用，解题的关键是：（1）找准等量关系，正确列出二元一次方程组；（2）分别求出到两商城购买所需费用． 28．（7分）我们规定：a﹣p＝
2
（a≠0），即a的负P次幂等于a的p次幂的倒数．例：4﹣
＝
；（﹣2）﹣2＝
；
，那么a＝ ±4 ；
（1）计算：5﹣2＝
（2）如果2﹣p＝，那么p＝ 3 ；如果a﹣2＝
（3）如果a﹣p＝，且a、p为整数，求满足条件的a、p的取值． 【分析】（1）根据负整数指数幂的计算法则计算即可求解； （2）根据负整数指数幂的计算法则找到指数即可求解；
（3）根据负整数指数幂的计算法则找到底数和指数即可求解． 解：（1）5﹣2＝
；（﹣2）﹣2＝；
，那么a＝±4；
（2）如果2﹣p＝，那么p＝3；如果a﹣2＝（3）由于a、p为整数， 所以当a＝9时，p＝1； 当a＝3时，p＝2； 当a＝﹣3时，p＝2． 故答案为：（1）；；（2）3；±4． 【点评】考查了负整数指数幂，负整数指数幂：a＝﹣p（a≠0，p为正整数），注意：①a﹣2
≠0；②计算负整数指数幂时，一定要根据负整数指数幂的意义计算，避免出现（﹣3）＝（﹣3）×（﹣2）的错误；③当底数是分数时，只要把分子、分母颠倒，负指数就可变为正指数；④在混合运算中，始终要注意运算的顺序． 免责声明：本文仅代表作者个人观点，作参考，并请自行核实相关内容. 声明：本文部分内容来自网络，本司不为其真实性负责，如有异议请及时联系，本司将予以删除 本文部分内容来自网络，本人不为其真实性负责，如有异议请及时联系，本人将予以删除BaidubaidubaidubaidubadiubadiuBaidubaidubaidubaidubadiubadiuBaidubaidubaidubaidubadiubadiuBaidubaidubaidubaidubadiubadiuBaidubaidubaidubaidubadiubadiuBaidubaidubaidubaidubadiubadiuBaidubaidubaidubaidubadiubadiuBaidubaidubaidubaidubadiubadiu BaidubaidubaidubaidubadiubadiuBaidubaidubaidubaidubadiubadiuBaidubaidubaidubaidubadiubadiuBaidubaidubaidubaidubadiubadiuBaidubaidubaidubaidubadiubadiuBaidubaidubaidubaidubadiubadiuBaidubaidubaidubaidubadiubadiuBaidubaidubaidubaidubadiubadiu BaidubaidubaidubaidubadiubadiuBaidubaidubaidubaidubadiubadiuBaidubaidubaidubaidubadiubadiuBaidubaidubaidubaidubadiubadiuBaidubaidubaidubaidubadiubadiuBaidubaidubaidubaidubadiubadiuBaidubaidubaidubaidubadiubadiuBaidubaidubaidubaidubadiubadiu BaidubaidubaidubaidubadiubadiuBaidubaidubaidubaidubadiubadiuBaidubaidubaidubaidubadiubadiuBaidubaidubaidubaidubadiubadiuBaidubaidubaidubaidubadiubadiuBaidubaidubaidubai
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北京市门头沟区2018-2019学年初一上册期末数学试卷
