一道初二数学题,急!设a=x/y+z,b=y/x+z,c=z/x+y,且x+y+z不等于0．求代数式a/a+1+b/b+1=c/c+1的值．要具体的过程!为什么因a=x/y+z,b=y/x+z,c=z/x+y所以a+1=(x+y+z)/y+z,b+1=(x+y+z)/x+z,c+1=(a+y+z)/x+y?看不明白啊.

Question

一道初二数学题,急!
设a=x/y+z,b=y/x+z,c=z/x+y,且x+y+z不等于0．求代数式a/a+1+b/b+1=c/c+1的值．
要具体的过程!
为什么因a=x/y+z,b=y/x+z,c=z/x+y 所以a+1=(x+y+z)/y+z,b+1=(x+y+z)/x+z,c+1=(a+y+z)/x+y ?看不明白啊.

Answer 1

因a=x/y+z,b=y/x+z,c=z/x+y
所以a+1=(x+y+z)/y+z,b+1=(x+y+z)/x+z,c+1=(a+y+z)/x+y
a/a+1+b/b+1+c/c+1
=(1-1/a+1)+(1-1/b+1)+(1-1/c+1)
=3-((1/a+1)+(1/b+1)+(1/c+1))
=3-(((y+z)/(x+y+z))+((x+z)/(x+y+z))+((x+y)/(x+y+z)))
=3-2((x+y+z)/(x+y+z))
=3-2
=1
说明:a/a+1=(a+1-1)/(a+1)=(a+1)/(a+!)-1/(a+1)=1-1/(a +1),其它类似.
有不清楚的地方发消息给我
用了等式两边都加上1.
a=x/y+z→a+1=(x/y+z)+1,b=y/x+z→b+1=(y/x+z)+1,
c=z/x+y →c+1=(z/x+y)+1.
a+1=(x/y+z)+1=(x/y+z)+(y+z)/(y+z)=(x+y+z)/(y+z)
这里1=(y+z)/(y+z),然后同分母的分式相加分母不变,分子相加.
其它两个等式方法同上