Question

初三数学急求！

若abc=1，x/(1+a+ab)+x/(1+b+bc)+x/(1+c+ac)=2003，则x=________

Answer 1

因为 ABC = 1，所以：
1 + B + BC = ABC + B + BC = B (1 + C + AC)
1 + A + AB = ABC + A + AB = A (1 + B + BC) = AB (1 + C + AC)
所以：
X/(1 + A + AB) + X/(1 + B + BC) + X/(1 + C + AC)
= X/[AB(1 + C + AC)] + X/[B(1 + C + AC)] + X/(1 + C + AC)
= [ X/(1 + C + AC) ] * (1/AB + 1/B + 1)
= [ X/(1 + C + AC) ] * [ (1 + A + AB)/AB ]
= [ X/(1 + C + AC) ] * [ AB (1 + C + AC) / AB ]
= X
= 2006
所以：X = 2006

Answer 2

X/(abc+a+ab)+X/(abc+b+bc)+X/(abc+c+bc)=(1/a+1/b+1/c)*X/(1+a+ab)+X/(1+b+bc)+X/(1+c+bc)=2003

x*(1/a+1/b+1/c)=1

x=1/(ab+ac+bc)

Answer 3

6009