家儿_kk
幼苗
共回答了15个问题采纳率:86.7% 举报
(1)f(x)+f(
1
x )=(12分)
f(x)+f(
1
x )=
x 2
1+ x 2 +
1
x 2
1+
1
x 2 =1(5分)
(2)
f(1)+f(2)+f(3)++f(2010)+f(
1
2 )+f(
1
3 )++f(
1
2010 )
=2009+
1
2 =
4019
2 (8分)
(3)设0<x 1 <x 2
f( x 1 )-f( x 2 )=
x 21
1+
x 21 -
x 22
1+
x 22 =
x 21 (1+
x 22 )-
x 22 (1+
x 21 )
(1+
x 21 )(1+
x 22 )
=
x 21 -
x 22
(1+
x 21 )(1+
x 22 )
=
( x 1 + x 2 )( x 1 - x 2 )
(1+
x 21 )(1+
x 22 ) (11分)
由0<x 1 <x 2 知x 1 -x 2 <0(12分)
所以有
( x 1 + x 2 )( x 1 - x 2 )
(1+
x 21 )(1+
x 22 ) <0 即f(x 1 )-f(x 2 )<0
所以f(x 1 )<f(x 2 )
函数 f(x)=
x 2
1+ x 2 在区间(0,+∞)上为增函数(14分)
1年前
7