试题:

已知=+, 且x1+x2<0, x2+x3<0, x3+x1<0则(    )
A f(x1)+f(x2)+f(x3)>0  B f(x1)+f(x2)+f(x3)<0  C f(x1)+f(x2)+f(x3)="0 " D f(x1)+f(x2)+f(x3)符号不能确定.

答案:

我来补答
B
=3+1,∴>0∴在上是增函数,且是奇函数,
∴f(x1)<f(-x2), f(x2)<f(-x3), f(x3)<f(-x1)∴f(x1)+f(x2)+f(x3)<-[f(x1)+f(x2)+f(x3)]即f(x1)+f(x2)+f(x3)<0故选B
 
 
展开全文阅读
剩余:2000
这些题目你会做吗?