有已知可以得到:a-b=-1,b-c=-1,c-a=2,c-b=1
原式=(a^2+b^2+c^2-bc-ac-ab)/abc
=【(a-b)^2+(b-c)^2+(a-c)^2]/abc
=(1+1+4)/6024
=1/1004
3x+5/(x+2)(x+3)=A/x+2+B/x+3
(3x+5)/(x+2)(x+3)=[(A+B)x+(3A+2B)]/(x+2)(x+3)
A+B=3
3A+2B=5
A=-1
B=4
(3A+B)^2008
=(-3+4)^2008
=1
有已知可以得到:a-b=-1,b-c=-1,c-a=2,c-b=1
原式=(a^2+b^2+c^2-bc-ac-ab)/abc
=【(a-b)^2+(b-c)^2+(a-c)^2]/abc
=(1+1+4)/6024
=1/1004
3x+5/(x+2)(x+3)=A/x+2+B/x+3
(3x+5)/(x+2)(x+3)=[(A+B)x+(3A+2B)]/(x+2)(x+3)
A+B=3
3A+2B=5
A=-1
B=4
(3A+B)^2008
=(-3+4)^2008
=1