bx+1 |
2x+a |
1 |
x |
b
| ||
2
|
b+x |
2+ax |
则f(x)f(
1 |
x |
bx+1 |
2x+a |
b+x |
2+ax |
则f(x)f(
1 |
x |
(bx+1)(b+x)−k(2x+a)(2+ax) |
(2x+a)(2+ax) |
=
(b−2ak)x2+(b2+1−4k−ka2)x+b−2ak |
(2x+a)(2+ax) |
则
|
4a2k2-(4+a2)k+1=0,
解得,k=
1 |
4 |
1 |
a2 |
若k=
1 |
a2 |
2 |
a |
与ab≠2相矛盾,舍去.
则k=
1 |
4 |
故答案为
1 |
4 |
bx+1 |
2x+a |
1 |
x |
bx+1 |
2x+a |
1 |
x |
b
| ||
2
|
b+x |
2+ax |
1 |
x |
bx+1 |
2x+a |
b+x |
2+ax |
1 |
x |
(bx+1)(b+x)−k(2x+a)(2+ax) |
(2x+a)(2+ax) |
(b−2ak)x2+(b2+1−4k−ka2)x+b−2ak |
(2x+a)(2+ax) |
|
1 |
4 |
1 |
a2 |
1 |
a2 |
2 |
a |
1 |
4 |
1 |
4 |