第1题 已知:abc≠0 a+b+c=0 证:a(1/b+1/c)+b(1/a+1/c)+c(1/b+1/c)+3=0第2题 设:abc=1 求a/(ab+a+1)+b/(bc+b+1)+c/(ca+c+1)的值
第1题 已知:abc≠0 a+b+c=0 证:a(1/b+1/c)+b(1/a+1/c)+c(1/b+1/c)+3=0
第2题 设:abc=1 求a/(ab+a+1)+b/(bc+b+1)+c/(ca+c+1)的值
1.证:因为abc≠0可知,a,b,c均不为0
又 a(1/b+1/c)+b(1/a+1/c)+c(1/b+1/a)+3
=a/b+a/c+b/a+b/c+c/b+c/a+b/b+a/a+c/c
=(a+b+c)/a+(a+b+c)/b+(a+b+c)/c
因a+b+c=0
所以上式结果为0
即a(1/b+1/c)+b(1/a+1/c)+c(1/b+1/c)+3=0
注:这里面的式子可能是你抄错了 ,不应该是c(1/b+1/c),而是c(1/b+1/a)。
a(1/b+1/c)+b(1/a+1/c)+c(1/b+1/a)+3
=a/b+a/c+b/a+b/c)+c/b+1/a+3
=(b+c)/a+(a+c)/b+(a+b)/c+3
a+b+c=0 b+c=-a a+c=-b a+b=-c
a(1/b+1/c)+b(1/a+1/c)+c(1/b+1/a)+3
=(b+c)/a+(a+c)/b+(a+b)/c+3
=-a/a+(-b)/b+(-c)/c+3
=0
a/(ab+a+1)+b/(bc+b+1)+c/(ca+c+1) abc=1
=a/(ab+a+1)+ab/(abc+ab+a)+cab/(ca*ab+cab+ab)
=a/(ab+a+1)+ab/(1+ab+a)+1/(a+1+ab)
=(ab+a+1)/(ab+a+1)
=1
第一题,a(1/b+1/c)+b(1/a+1/c)+c(1/b+1/a)+3=0 ,已知abc≠0,所以可以在等式左右同时乘abc,再整理得到ac(a+b+c)+ab(a+b+c)+bc( a+b+c)=0 ,已知( a+b+c)=0,所以等式成立。
遗撼有人比我快,我就不答了
(1)将a=-b-c带入到等式左边
a(1/b+1/c)+b(1/a+1/c)+c(1/b+1/c)+3
=(-b-c)(1/b+1/c)+b(1/a+1/c)+c(1/b+1/a)+3
= -2-b/c-c/b+b/a+b/c+c/b+c/a+3
= (b+c)/a +1
= 0
(2)
a/(ab+a+1)
= a/(1/c+a+1)
= ac/(ac+c+1)
b/(bc+b+1)
=b/(1/a+b+1)
=ab/(ab+a+1)
=ab/(1/c+a+1)
=abc/(ac+c+1)
=1/(ac+a+1)
所以原式=ac/(ac+c+1)+1/(ac+a+1)+c/(ca+c+1)
=1
1、因为证明条件长,不妨先化简,所以用了分析法须证:a(1/b+1/c)+b(1/a+1/c)+c(1/b+1/c)+3=0即证:a(1/b+1/c)+b(1/a+1/c)+c(1/a+1/b)=-3下面进行变形:a(1/a+1/b+1/c)+b(1/a+1/b+1/c)+c(1/a+1/b+1/c)-3=-3(左边加上a...