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展开全部设a/（b+c）=ka=k（b+c），同理b=k（a+c）c=k（a+b）三式相加a+b+c=2k（a+b+c）k=1/2c/（a+b）=1/2a+b=2c代入2a+2b+c/a+b-3c=（4c+c）/（2c-3c）=5c/（-c）=-5
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人教版八年级数学题。1/b指b分之1，其它类推abcd是互不相等的四个数...
人教版八年级数学题。1/b 指b分之1，其它类推a b c d 是互不相等的四个数
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我可以负责的说，题目少条件，如四个数都为1，则x为2，四个数都为2，则x为2.5分析：等式看看为4个方程，却有5个未知数，显然不可解，则5个量必有某种关系。 通过解a+1/b=b+1/c=c+1/d=d+1/a，(把a当做未知之数）得b=c=d=a，x=a+1/a其实这是个轮换式，如果了解它得性质定理（一般竞赛书都有），很容易得abcd四个数相等，那么可得x已赞过已踩过你对这个回答的评价是？评论
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呵呵，三楼你看看这个成立么？令x=0，a=1则b=-1c=1d=-1abcd相等吗？不会别瞎蒙，搞深沉！还负责的说！四楼你应用等比例定理的时候注意条件：分母不能为零！将上述结果带入你答案显然不对嘛。。。一楼二楼不说了，难过！
本回答被提问者采纳x=[ab+bc+cd+da+4]/a+b+c+d.过程是，由a+1/b=x得ab+1=bx,同理有bc+1=cx,cd+1=dx,da+1=ax,四式相加得，ab+bc+cd+da+4=x(a+b+c+d),解得x=(ab+bc+cd+da+4)/a+b+c+d
a+1/b=(a+b)/b=x
a+b=bx同理可以得到：
b+c=cx
c+d=dx
a+d=ax四个式子相加，得：a+b+b+c+c+d+a+d=ax+bx+cx+dx即2(a+b+c+d)=(a+b+c+d)xx=2

过程说明下怎么算?...
过程说明下怎么算?
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令a/b=b/c=c/a=k则a=kbb=kcc=ka相加a+b+c=k(a+b+c)(a+b+c)(k-1)=0a+b+c=0或k-1=0若a+b+c=0,则(a+b+c)/c=0若k-1=0,k=1,则a=b=c,则(a+b+c)/c=3c/c=3所以(a+b+c)/c=0或3
本回答由提问者推荐已赞过已踩过你对这个回答的评价是？评论
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a/b=b/c=c/a=(a+b+c)/(a+b+c)=1a=b=c(a+b+c)/c=3.