a b c⒈已知abc=1,求_______+_________+_________的值.ab+a+1 bc+b+1 a

1.因为：abc=1
a/ab+a+1= a/ab+a+abc= 1/bc+b+1
c/ca+c+1= c/(1/b)+c+1= bc/bc+b+1
所以：
原式
= a/(ab+a+1)+ b/(bc+b+1)+ c/(ca+c+1)
= 1/(bc+b+1)+ b/(bc+b+1)+ bc/(bc+b+1)
= bc+b+1/bc+b+1
= 1
2.(a-b)/(2a+2b)-(a^2+b^2)/(a^2-b^2)
=(a-b)²/[2(a-b)(a+b)]-2(a²+b²)/[2(a-b)(a+b)]
=-(a+b)²/[2(a-b)(a+b)]
=-(a+b)/[2(a-b)]
3.已知(2x-3)/(x-1)(x+2)=A/(x-1)+B/(x+2),求A、B的值.
A/(x-1)+B/(x+2)
=A(x+2)/(x+2)(x-1)+B(x-1)/(x+2)(x-1)
=[A(x+2)+B(x-1)]/(x+2)(x-1)
=[(A+B)x+(2A-B)]/(x+2)(x-1)=2x-3/[(x-1)(x+2)]
A+B=2,2A-B=-3
3A=-1
A=-1/3
B=2+1/3=7/3

确实有些乱，重写一下吧！！

1。不妨令a=b=c=1,代入a/(ab+a+1)+b/(bc+b+1)+c/(ac+c+1)=1/3+1/3+1/3=1
2。(a-b)/(2a+2b)-(a^2+b^2)/(a^2-b^2)
=(a-b)²/[2(a-b)(a+b)]-2(a²+b²)/[2(a-b)(a+b)]
=-(a+b)²/[2(a-b)(a+b)]

这是什么意思？