解:(1)设药物燃烧阶段函数解析式为.由题意得: ························································································································ 2分 .此阶段函数解析式为······································································· 3分 (2)设药物燃烧结束后的函数解析式为.由题意得: ·························································································································· 5分 .此阶段函数解析式为······································································ 6分 (3)当时.得···················································································· 7分 ························································································································· 8分 ·························································································································· 9分 从消毒开始经过50分钟后学生才可回教室.···························································· 10分 查看更多

 

题目列表(包括答案和解析)

方程
x+2
x
+
x
3x+6
=2,用换元法解,若设
x+2
x
=y,则此方程化为整式方程的是(  )

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七年级学生参加了社会实践调查活动,到生态果园调查后得到如下信息:今年收获了15吨李子和8吨桃子,要租用甲、乙两种货车共6辆,及时运往外地,经询问,甲种货车可装李子4吨和桃子1吨,乙种货车可装李子1吨和桃子3吨.根据同学们带回的信息,试探究以下问题:
(1)共有几种租车方案?
(2)经咨询运输公司,甲种货车每辆需付运费1000元,乙种货车每辆需付运费700元,试帮助选出最佳方案,并求出此方案运费是多少.
请同学们补充完成下列部分解题过程:
(1)解:
①若设租用甲车x辆,则租用乙车
(6-x)
(6-x)
辆,
②由题意可知:甲车一共可装
x
x
吨桃子,乙车一共可装
3(6-x)
3(6-x)
吨桃子,则甲,乙两种车一共可装
x+3(6-x)
x+3(6-x)
吨桃子.(用含有x的代数式表示)
请列出不等式
x+3(6-x)≥8
x+3(6-x)≥8

③甲车一共可装
4x
4x
吨李子,乙车一共可装
(6-x)
(6-x)
吨李子,则甲,乙两种车一共可装
4x+(6-x)
4x+(6-x)
吨李子.(用含有x的代数式表示)
请列出不等式
4x+(6-x)≥15
4x+(6-x)≥15

④请列出不等式组,并求出满足不等组的整数解,写出相应的方案
(2)解:

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25、已知:两个正整数的和与积相等,求这两个正整数.
解:不妨设这两个正整数为a、b,且a≤b.
由题意,得ab=a+b,(*)
则ab=a+b≤b+b=2b,所以a≤2,
因为a为正整数,所以a=1或2,
①当a=1时,代入等式(*),得1•b=1+b,b不存在;
②当a=2时,代入等式(*),得2•b=2+b,b=2.
所以这两个正整数为2和2.
仔细阅读以上材料,根据阅读材料的启示,思考是否存在三个正整数,它们的和与积相等试说明你的理由.

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用换元法解方程,若设,则原方程化为关于的整式方程是

A、            B、       

C、                D、

 

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用换元法解方程,设,那么原方程可化为   

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