题目
题型:松江区二模难度:来源:
OP |
OQ |
OA |
OB |
OA1 |
OA2 |
OA3 |
OA4 |
OA5 |
OA |
OB |
答案
类比点P,Q是线段AB的三等分点,则有
OP |
OQ |
OA |
OB |
得:
OA1 |
OA5 |
OA2 |
OA4 |
OA3 |
OA |
OB |
所以
OA1 |
OA2 |
OA3 |
OA4 |
OA5 |
5 |
2 |
OA |
OB |
故答案为
5 |
2 |
核心考点
试题【如图,有以下命题成立:设点P,Q是线段AB的三等分点,则有OP+OQ=OA+OB.将此命题推广,设点A1,A2,A3,A4,A5是线段AB的六等分点,则OA1+】;主要考察你对合情推理与演译推理等知识点的理解。[详细]
举一反三
a |
x |
a |
a |
b |
x2 |
4 | b |
4 | b |
c |
xn |
a+b |
2 |
A.① | B.② | C.①② | D.③ |