初二数学分式题。

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.
∵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

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