Question

（1）已知abc=1，求a/(ab+a+1）+b/(bc+b+1)+c/(ac+c+1）的值

（1）已知abc=1，求a/(ab+a+1）+b/(bc+b+1)+c/(ac+c+1）的值（提示:可以用分式的基本性质把异分母加法转化为同分母加法，如：a/（ab+a+1)=(a*c)/[c（ab+a+1)]=ac/(1+ac+c)）；
（2）已知a+b+c=0,求a(1/b+1/c)+b(1/a+1/c）+c(1/a+1/b)的值

还要过程

Answer 1

（1）因为abc=1.
a/(ab+a+1)+b/(bc+b+1)+c/(ac+c+1)
=a/(ab+a+1)+ab/(abc+ab+a)+abc/(abac+abc+ab)
=a/(ab+a+1)+ab/(1+ab+a)+1/(a+1+ab)
=(ab+a+1)/(ab+a+1)
=1
(2)a﹙1/b＋1/c﹚＋b﹙1/a＋1/c﹚＋c﹙1/a＋1/b﹚
＝a﹙b＋c﹚/bc＋b﹙a＋c﹚/ac＋c﹙a＋b﹚/ab
＝﹣a²/bc－b²/ac－c²/ab
＝﹣﹙a³＋b³＋c³﹚/abc
＝﹣[a³＋b³－﹙a＋b﹚³]/abc
＝﹣[a³＋b³－a³－b³－3a²b－3ab²]/abc
＝3ab﹙a＋b﹚/abc
＝3﹙a＋b﹚/c
＝﹣3c/c
＝﹣3.

Answer 2

a/(ab+a+1)
=a/(ab+a+abc)
=1/(bc+b+1)

a/(ab+a+1)
=(ac)/(abc+ac+c)
=(ac)/(ca+c+1)

进行类似变换可得

3[a/(ab+a+1)+b/(bc+b+1)+c/(ca+c+1)]
=a/(ab+a+1)+1/(bc+b+1)+(ac)/(ca+c+1)+b/(bc+b+1)+1/(ca+c+1)+(ab)/(ab+a+1)+c/(ca+c+1)+1/(ab+a+1)+(bc)/(bc+b+1)
=(ab+a+1)/(ab+a+1)+(bc+b+1)/(bc+b+1)+(ca+c+1)/(ca+c+1)
=1+1+1
=3

a/(ab+a+1)+b/(bc+b+1)+c/(ca+c+1)=1

Answer 3

﹣3