题目内容
已知函数f(x)=
(x∈R),若x1+x2=1,则f(x1)+f(x2)=______;若n∈N*,则f(
)+f(
)+…+f(
)+f(
)=______.
| 1 |
| 4x+2 |
| 1 |
| n |
| 2 |
| n |
| n-1 |
| n |
| n |
| n |
∵函数f(x)=
(x∈R),
∴f(x1)=
又∵x1+x2=1,x2=1-x1,
∴f(x2)=
f(x1)+f(x2)=
+
=
+
=
=
∴f(
)+f(
)+…+f(
)+f(
)
=[f(
)+f(
)]+[f(
)+f(
)]+…+f(
)
=
•
+f(1)
=
-
故答案为:
;
-
.
| 1 |
| 4x+2 |
∴f(x1)=
| 1 |
| 4x1+2 |
又∵x1+x2=1,x2=1-x1,
∴f(x2)=
| 1 |
| 4(1-x1)+2 |
f(x1)+f(x2)=
| 1 |
| 4x1+2 |
| 1 |
| 4(1-x1)+2 |
| 2 |
| 2•4x1+4 |
| 4x1 |
| 4 +2•4x1 |
| 2+4x1 |
| 4 +2•4x1 |
| 1 |
| 2 |
∴f(
| 1 |
| n |
| 2 |
| n |
| n-1 |
| n |
| n |
| n |
=[f(
| 1 |
| n |
| n-1 |
| n |
| 2 |
| n |
| n-1 |
| n |
| n |
| n |
=
| n-1 |
| 2 |
| 1 |
| 2 |
=
| n |
| 4 |
| 1 |
| 12 |
故答案为:
| 1 |
| 2 |
| n |
| 4 |
| 1 |
| 12 |
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