初二数学分式题。
若abc=1，求证1/（ab+a+1)+1/(bc+b+1)+1/(ca+c+1)=1.
已知(1/x)-(1/y)=2001,求分式（x-xy-y)/(x-y)的值。
这 。。。 我还是发起投票把。 不过我感谢第一位。
1/（ab+a+1)+1/(bc+b+1)+1/(ca+c+1)
=1/（1/c+a+1)+1/(bc+b+1)+1/(ca+c+1).
=c/（1+ac+c)+1/(bc+b+1)+1/(ca+c+1)
=c+1/（1+ac+c)+1/(bc+b+1)
=c+1/（1+1/b+c)+1/(bc+b+1)
=b(c+1)/（b+1+bc)+1/(bc+b+1)
=(bc+b+1)/（b+1+bc)
=1
（x-xy-y)/(x-y)
=1-xy/(x-y)
=1-1/(1/y-1/x)
=1+1/(1/x-1/y)
=1+1/2001
=2002/2001
=1/（1/c+a+1)+1/(bc+b+1)+1/(ca+c+1).
=c/（1+ac+c)+1/(bc+b+1)+1/(ca+c+1)
=c+1/（1+ac+c)+1/(bc+b+1)
=c+1/（1+1/b+c)+1/(bc+b+1)
=b(c+1)/（b+1+bc)+1/(bc+b+1)
=(bc+b+1)/（b+1+bc)
=1
（x-xy-y)/(x-y)
=1-xy/(x-y)
=1-1/(1/y-1/x)
=1+1/(1/x-1/y)
=1+1/2001
=2002/2001
温馨提示：内容为网友见解，仅供参考
第1个回答 2011-08-23
1.
∵abc=1,∴ab=1/c,ac=1/b
左边=1/(1/c+ac/c+c/c)+1/(bc+b+1)+1/(ca+c+1)
=(c+1)/(ac+c+1)+1/(bc+b+1)
=(c+1)/[(bc+b+1)/b]+1/(bc+b+1)
=(bc+b+1)/(bc+b+1)
=1
∴左边=右边
∴1/（ab+a+1)+1/(bc+b+1)+1/(ca+c+1)=1.
2.
∵(1/x)-(1/y)=2001
∴(y-x)/(xy)=2001,(xy)/(y-x)=1/2001
（x-xy-y)/(x-y)
=[(x-y)-xy]/(x-y])
=1-(xy)/(x-y)
=1+(xy)/(y-x)
=1+1/2001
=2002/2001
∴分式（x-xy-y)/(x-y)的值为2002/2001
∵abc=1,∴ab=1/c,ac=1/b
左边=1/(1/c+ac/c+c/c)+1/(bc+b+1)+1/(ca+c+1)
=(c+1)/(ac+c+1)+1/(bc+b+1)
=(c+1)/[(bc+b+1)/b]+1/(bc+b+1)
=(bc+b+1)/(bc+b+1)
=1
∴左边=右边
∴1/（ab+a+1)+1/(bc+b+1)+1/(ca+c+1)=1.
2.
∵(1/x)-(1/y)=2001
∴(y-x)/(xy)=2001,(xy)/(y-x)=1/2001
（x-xy-y)/(x-y)
=[(x-y)-xy]/(x-y])
=1-(xy)/(x-y)
=1+(xy)/(y-x)
=1+1/2001
=2002/2001
∴分式（x-xy-y)/(x-y)的值为2002/2001
第2个回答 2011-08-23
你是找题目？追问
不好意思，提问时写不下了，补充好了。
追答第2题 答案2000/2001
追问不好意思，我想要过程，您麻烦一下好吗？真是谢谢了，我会追加的、
追答(1/x)-(1/y)=2001
合并的(Y-X)/XY=2001
（x-xy-y)/(x-y)=（X-Y）/(X-Y)-XY/(X-Y)=1-XY/(X-Y)=1-1/2001=2000/2001
真是太谢谢了，对头，第一题您有头绪吗？能帮忙答一下吗？
追答第一题解起来太复杂，按照第二题的思路做就可以了
相似回答
