题目内容
给出下列四个结论:
①已知△ABC中,三边a,b,c满足(a+b+c)(a+b-c)=3ab,则∠C等于120°.
②若等差数列an的前n项和为Sn,则三点(10,
),(100,
),(110,
)共线.
③等差数列an中,若S10=30,S20=100,则S30=210.
④设f(x)=
,则f(-8)+f(-7)+…+f(0)+…+f(8)+f(9)的值为
.
其中,结论正确的是 ______.(将所有正确结论的序号都写上)
①已知△ABC中,三边a,b,c满足(a+b+c)(a+b-c)=3ab,则∠C等于120°.
②若等差数列an的前n项和为Sn,则三点(10,
| S10 |
| 10 |
| S100 |
| 100 |
| S110 |
| 110 |
③等差数列an中,若S10=30,S20=100,则S30=210.
④设f(x)=
| 1 | ||
2x+
|
9
| ||
| 2 |
其中,结论正确的是 ______.(将所有正确结论的序号都写上)
①由(a+b+c)(a+b-c)=3ab,得到(a+b)2-c2=3ab,化简得:a2+b2-c2=ab,
则cosC=
=
=
,根据C∈(0,180°),得到∠C=60°,所以此选项错误;
②因为
=
=a1+
d,同理
=a1+
d,
=a1+
d,
则
=
=
=
=
=
,
所以三点(10,
),(100,
),(110,
)共线.此选项正确;
③根据等差数列的性质可知,S10,S20-S10,S30-S20成等差数列,
得到:2(S20-S10)=S10+(S30-S20),将S10=30,S20=100,
代入得:2(100-30)=30+(S30-100),解得:S30=210.此选项正确;
④因为f(x)+f(1-x)=
+
=
+
=
+
=
=
=
,
则f(-8)+f(-7)+…+f(0)+…+f(8)+f(9)=
×9=
.此选项正确.
所以,正确的结论序号有:②③④.
故答案为:②③④
则cosC=
| a2+b2-c2 |
| 2ab |
| ab |
| 2ab |
| 1 |
| 2 |
②因为
| S10 |
| 10 |
10a1+
| ||
| 10 |
| 9 |
| 2 |
| S100 |
| 100 |
| 99 |
| 2 |
| S110 |
| 110 |
| 109 |
| 2 |
则
| ||||
| 100-10 |
(a1+
| ||||
| 90 |
| d |
| 2 |
| ||||
| 110-100 |
(a1+
| ||||
| 10 |
| d |
| 2 |
所以三点(10,
| S10 |
| 10 |
| S100 |
| 100 |
| S110 |
| 110 |
③根据等差数列的性质可知,S10,S20-S10,S30-S20成等差数列,
得到:2(S20-S10)=S10+(S30-S20),将S10=30,S20=100,
代入得:2(100-30)=30+(S30-100),解得:S30=210.此选项正确;
④因为f(x)+f(1-x)=
| 1 | ||
2x+
|
| 1 | ||
21-x+
|
=
| 1 | ||
2x+
|
| 2x | ||
2 +
|
| ||||
|
| 2x | ||
2+
|
=
| ||
1 +
|
| ||||
|
| ||
| 2 |
则f(-8)+f(-7)+…+f(0)+…+f(8)+f(9)=
| ||
| 2 |
9
| ||
| 2 |
所以,正确的结论序号有:②③④.
故答案为:②③④
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