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列一元一次方程解应用题(复习课)教学目标:1、会借助示意图、表格分析题意,寻找相等关系;2、通过列方程解应用题,培养学生分析问题、解决问题的能力;5、检查所得的值是否正确和符合实际情形,并写出答案(包括单位名称)。1、弄清题意,用字母(如X)表示问题里的未知数;2、分析题意,找出相等关系(可借助于示意图、表格);3、根据相等关系,列出需要的代数式,从而列出方程;(注意:左右两边单位统一,已知条件都要用上)4、解这个方程,求出未知数的值;填空:1、甲乙两车从相距108千米处同时出发,相向而行,已知甲车速度为19千米/时,乙车速度为17千米/时———小时后,两车在相遇前相距36千米。22、(接上题)又过了———小时,两车相遇后相距36千米。由相遇前相距36千米,到相遇后相距36千米,两车共行驶———千米。23、若甲乙两车从相距108千米A、B处同时相向而行,已知甲车速度为19千米/时,乙车速度为17千米/时,到达B、A后立即返回原处,两车第一次相遇时间为———小时,从第一次相遇到第二次相遇的时间为———小时。6372例1:甲、乙两车从A、B两地于上午8点钟同时出发,相向而行,已知甲的速度比乙快2千米/时,到上午10时两车还相距36千米,又过了两小时后,两车又相距36千米。1、求甲乙两地间的距离与两车的速度;2、若甲乙两车分别从A、B两地同时相向而行,到B、A两地后立即返回,求两车第一次相遇和第二次相遇所走的时间是多少?分析:甲乙→←分析:甲乙→→←←分析:甲乙→→→←←←分析:甲乙→→→→←←←←分析:甲乙→→→→→←←←←分析:甲乙→→→→→→←←←←←分析:甲乙→→→→→→→←←←←←←分析:甲乙→→→→→→→→←←←←←←←分析:甲乙→→→→→→→→→←←←←←←←分析:甲乙→→→→→→→→→→←←←←←←←←分析:甲乙→→→→→→→→→→→←←←←←←←←←分析:甲乙→→→→→→→→→→→→←←←←←←←←←←分析:甲乙→→→→→→→→→→→→←←←←←←←←←←←分析:甲乙→→→→→→→→→→→→→←←←←←←←←←←←←36千米AB甲乙甲行2小时的路程(S1)乙行2小时的路程(s2)甲乙36千米分析:甲乙→→→→→→→→→→→→→→←←←←←←←←←←←←←分析:甲乙→→→→→→→→→→→→→→→←←←←←←←←←←←←←←分析:甲乙→→→→→→→→→→→→→→→→←←←←←←←←←←←←←←←分析:甲乙→→→→→→→→→→→→→→→→→←←←←←←←←←←←←←←←←分析:甲乙→→→→→→→→→→→→→→→→→→→→→→→→→→→→→→→→→←←←←←←←←←←←←←←←←←←分析:甲乙→→→→→→→→→→→→→→→→→→→→→→→→→→→→→→→→→→←←←←←←←←←←←←←←←←←←分析:甲乙→→→→→→→→→→→→→→→→→→→→→→→→→→→→→→→→→→→←←←←←←←←←←←←←←←←←←←←分析:甲乙→→→→→→→→→→→→→→→→→→→→→→→→→→→→→→→→→→→←←←←←←←←←←←←←←←←←←←←←分析:甲乙→→→→→→→→→→→→→→→→→→→→→→→→→→→→→→→→→→→←←←←←←←←←←←←←←←←←←←←←←分析:甲乙→→→→→→→→→→→→→→→→→→→→→→→→→→→→→→→→→→→→→←←←←←←←←←←←←←←←←←←←←←←分析:甲乙→→→→→→→→→→→→→→→→→→→→→→→←←←←←←←←←←←←←←←←←←←←←←←分析:甲乙→→→→→→→→→→→→→→→→→→→→→→→→←←←←←←←←←←←←←←←←←←←←←←←←分析:甲乙→→→→→→→→→→→→→→→→→→→→→→→→→←←←←←←←←←←←←←←←←←←←←←←←←←AB甲行2小时的路程乙行2小时的路程36千米36千米AB甲行2小时的路程(S1)乙行2小时的路程(s2)甲乙36千米甲乙解:⑴设乙车速度为X千米/时,则甲车速度为(X+2)千米/时。依题意列方程:2X+2(X+2)=72解得X=17,X+2=19,A、B两地距离为:72+36=108答:A、B两地距离是108千米,甲车速度为19千米/时,乙车速度为17千米/时。相等关系:前2小时所行驶的路程=后2小时所行驶的路程分析:甲乙→→→→→→→→→→→→→→→→→→→→→→→→→→←←←←←←←←←←←←←←←←←←←←←←←←←←分析:甲乙→→→→→→→→→→→→→→→→→→→→→→→→→→→←←←←←←←←←←←←←←←←←←←←←←←←←←←分析:甲乙→→→→→→→→→→→→→→→→→→→→→→→→→→→→→→←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←分析:甲乙→→→→→→→→→→→→→→→→→→→→→→→→→→→→→→→→←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←分析:甲乙→→→→→→→→→→→→→→→→→→→→→→→→→→→→→→→→→→→←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←分析:甲乙→→→→→→→→→→→→→→→→→→→→→→→→→→→→→→→→→→→→→←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←分析:甲乙→→→→→→→→→→→→→→→→→→→→→→→→→→→→→→→→→→→→→←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←分析:甲乙←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←→→→→→→→→→→→→→→→→→→→→→→→→→→→→→→→→→→→→→分析:甲乙←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←→→→→→→→→→→→→→→→→→→→→→→→→→→→→→→→→→→→→→←←←分析:甲乙←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←→→→→→→→→→→→→→→→→→→→→→→→→→→→→→→→→→→→→→←←←←←分析:甲乙←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←→→→→→→→→→→→→→→→→→→→→→→→→→→→→→→→→→→→→→←←←←←分析:甲乙→→→→→→→→→→→→→→→→→→→→→→→→→→→→→→→→→→→→→←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←分析:甲乙→→→→→→→→→→→→→→→→→→→→→→→→→→→→→→→→→→→→→←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←分析:甲乙→→→→→→→→→→→→→→→→→→→→→→→→→→→→→→→→→→→→→←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←分析:甲乙→→→→→→→→→→→→→→→→→→→→→→→→→→→→→→→→→→→→→←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←分析:甲乙→→→→→→→→→→→→→→→→→→→→→→→→→→→→→→→→→→→→→←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←→→分析:甲乙→→→→→→→→→→→→→→→→→→→→→→→→→→→→→→→→→→→→→←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←→→→分析:甲乙→→→→→→→→→→→→→→→→→→→→→→→→→→→→→→→→→→→→→←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←→→→→→分析:甲乙→→→→→→→→→→→→→→→→→→→→→→→→→→→→→→→→→→→→→←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←→→→→→→←←←←←←←←←←←←←←←←←←←←分析:甲乙→→→→→→→→→→→→→→→→→→→→→→→→→→→→→→→→→→→→→←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←→→→→→→→→→分析:甲乙→→→→→→→→→→→→→→→→→→→→→→→→→→→→→→→→→→→→→←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←→→→→→→→→→→→分析:甲乙→→→→→→→→→→→→→→→→→→→→→→→→→→→→→→→→→→→→→←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←→→→→→→→→→→→→→分析:甲乙→→→→→→→→→→→→→→→→→→→→→→→→→→→→→→→→→→→→→←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←←→→→→→→→→→→→→→→AB第二次相遇甲所行驶的路程第二次相遇乙所行驶的路程由以上分析可知:第一次相遇两车共行驶的路程为1个AB,到第二次相遇两车共行驶的路程为3个AB。⑵设第一次相遇为y小时19y+17y=108解得y=3设第二次相遇为a小时19a+17a=108×3解得a=9答:第一次相遇所行驶的时间为3小时,第二次相遇所行驶的时间为9小时。注意:一题中的几个小题,前题的结论可作后题的已知条件。相等关系:第一次相遇两车行驶路程和=108千米第二次相遇两车行驶路程和=108×3千米1、若两车相向而行,问何时两车相距36千米?(有两解)2、若两车在72千米的环形公路上,同时、同地、反向而行,甲车速19千米/时,乙车速17千米/时,问两车经过多少时间相遇?3、若两车在72千米的环形公路上,同时、同地、同向而行,甲车速19千米/时,乙车速17千米/时,当它们第一次相遇时需要多少时间?练习(辅助教材P57B组2、A组8):1、某学生总是以每小时5千米的速度行走,可以及时从家里走到学校,有一次他走了全程的三分之一后,搭上速度是每小时20千米的汽车,因此比原来提前2小时到校,他家离学校多远?分析:画示意图家学校步行:5千米/时家学校步行:5千米/时乘车:20千米/时相等关系:1、学生单独走的路程=学生步行和乘车所行路程和;2、学生单独走的时间-2小时=学生单独走三分之一路程的时间+乘车的时间;解法一:设学生单独走准时到校所用时间为t小时。5t=5×——+20(t–——–2)tt33解得t=4,5×4=20答:他家离学校20千米。解法二:设两地相距S千米。——–2=–——+—–—S3531——S32——S20解得:S=20答:两地相距20千米。2、有700克含碘15%的碘酒(碘溶解在酒精里就成为碘酒),应加入多少克纯酒精,才能得到含碘2%的碘酒?分析:原来的碘酒加纯酒精后,质量和浓度都发生了变化,但所含碘的质量没有变化,如图所示:15%的碘酒700克加纯酒精2%的碘酒(700+X)克X克相等关系:稀释含碘量不变加酒精前碘的质量=加酒精后碘的质量加酒精前加酒精后碘酒质量含碘浓度含碘质量700克(700+X)克15%2%700×15%克(700+X)2%克解:设需加酒精X克。700×15%=(700+X)2%解得X=4550答:需加纯酒精4550克。突出运用图示、表格分析等方法寻找相等关系。作业:辅助教材P57.B.2、3、4P58.B.10
本文标题:列一元一次方程解应用题(复习课)
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