1、ABC三个数，AB的最大公约数是15，BC的最大公约数是15，ABC的最大公约数是多少？2、一块正方形地，它的边长是20米，地中歌有横竖两条路，路宽2米，这块地可耕面...
1、ABC三个数，AB的最大公约数是15，BC的最大公约数是15，ABC的最大公约数是多少？2、一块正方形地，它的边长是20米，地中歌有横竖两条路，路宽2米，这块地可耕面积占全块地面积的几分之几？3、甲乙丙三个工程队歌修了一条路，甲乙共修了这条路的十二分之七，乙丙共修了这条路的四分之三，乙队修了这条路的几分之几？4、王叔叔从家到单位上班，如果每分钟走60米，就要迟到2分钟，如果每分钟走80米，就要早到3分钟，王叔叔从家到单位有多少米？5、甲乙两家超市村大米的吨数相等，甲超市售出大米3000吨，乙超市购入大米2000吨，这时乙超市的大米吨数是甲超市的2倍，甲乙两家超市原来共存大米多少吨？请各位帮帮忙啦！！各位，帮忙把步骤写出来哦！不要只写最后答案哦！(是小学五年级滴题。。。)
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展开全部1>解:假设15不是ABC的最大公约数设其最大公约数为X 又15是ABC的公约数所以X>15X为ABC的公约数则也为AB的公约数 又X>15与AB的最大公约数是15矛盾所以ABC的最大公约数是152>解:路的面积S=2*(2*20)-2*2=76所以可耕面积S2=324可耕面积占全块地面积=324/400=81%3>甲乙共修了这条路的十二分之七，乙丙共修了这条路的四分之三 所以设甲乙丙共修了这条路X则1+X=7/12+3/4所以X=1/34>解:设他恰好走X分钟不迟到60*(X+2)=80*(X-3)所以X=18故，王叔叔从家到单位有1200m5>解:设原来两家有米各X吨(X-3000)*2=X+2000所以X=8000t
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8000t还满意吧？
展开全部1 15 2 81/100 81% 3 1/3 4 1200m 5 8000t
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如图所示，在等腰三角形ABC中，一腰AC上的中线BM把△ABC的周长分为12cm和15cm两部分，求△ABC各边的长。...
如图所示，在等腰三角形ABC中，一腰AC上的中线BM把△ABC的周长分为12cm和15cm两部分，求△ABC各边的长。
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因为题目只说中线BM把△ABC周长分为12cm与15cm，所以要分两种情况讨论1.解：AM+AB=12①，BC+MC=15②，①+②得：AM+AB+BC+MC=27，∵AB=AC，AM+MC=AC，∴原式整理得：2AB+BC=27③，②-①得：BC+MC-AM-AB=3，∵AM=MC（中线定理）∴原式整理得：BC-AB=3④，将③④组成一个二元一次方程组，解得：AB=AC=8，BC=112.解：AM+AB=15①，BC+MC=12②，①+②得：AM+AB+BC+MC=27，整理得：2AB+BC=27③，①-②整理得：AB-BC=3④，将③④组成一个方程组，解得：AB=AC=10，BC=7
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Y=11所以
腰长为8，底长为16
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