已知aba+b=115,bcb+c=117,cac+a=116,则abcab+bc+ca的值是( )A. 121B. 122C. 123D. 124
问题描述:
已知
=ab a+b
,1 15
=bc b+c
,1 17
=ca c+a
,则1 16
的值是( )abc ab+bc+ca
A.
1 21
B.
1 22
C.
1 23
D.
1 24
答
∵
=ab a+b
,∴1 15
+1 a
=15①,1 b
∵
=bc b+c
,∴1 17
+1 b
=17②;1 c
∵
=ca c+a
,∴1 16
+1 a
=16③,1 c
∴①+②+③得,2(
+1 a
+1 b
)=48,1 c
∴
+1 a
+1 b
=24,1 c
则
=abc ab+bc+ca
=1
ab+bc+ac abc
=1
+1 c
+1 a
1 b
,1 24
故选D.
答案解析:先将上面三式相加,求出
+1 a
,1 b
+1 b
,1 c
+1 a
,再将1 c
化简即可得出结果.abc ab+bc+ca
考试点:对称式和轮换对称式.
知识点:本题考查了对称式和轮换对称式,是基础知识要熟练掌握.