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Vì \(M\)trung điểm của \(AB\Rightarrow AM=BM\)
\(OA+OB=OB+OB+AB=2OB+BM+BM=2OB+2BM\)(1)
\(OM=OB+BM\) (2)
Từ (1) và (2) \(\Rightarrow\frac{OA+AB}{OM}=\frac{2OB+2BM}{OB+BM}=\frac{2\left(OB+BM\right)}{OB+BM}=2\)
Mặt khác \(OM=\frac{OA+OB}{2}\)
A B M
\(OM< \frac{OA+OB}{2}\) KO BIẾT LÝ LUẬN NHƯNG BIẾT SO SÁNH
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Ta có:
2OM=OM+OM=(OA+AM)+(OB−BM)=OA+OB+(AM−BM)2OM=OM+OM=(OA+AM)+(OB−BM)=OA+OB+(AM−BM)
Vì M là trung điểm AB
=> AM=BM
=> 2OM=OA+OB2OM=OA+OB
=> OM=OA+OB2
Bài 1 :
a) Do O thuộc đoạn thẳng AM nên O nằm giữa hai điểm A và M .Ta có :\(OA< MA\)
M là trung điểm của AB nên M nằm giữa A và B và;
\(MA=MB=\frac{1}{2}AB\)
\(\Rightarrow MA< AB\)
\(\Rightarrow OA< MA< AB\) chứng tỏ M nằm giữa O và B
Do đó : \(OM=OB-MB\)
Mặt khác ,theo trên : O nằm giưa A và M nên \(OM=MA-OA\)
\(\Rightarrow20M=OB-OA\)( Vì \(MA=MB\))
\(\Rightarrow OM=\frac{1}{2}\left(OB-OA\right)\)
b) TRƯỜNG HỢP 2 :
O thuộc tia đối của AB
Do M là trung điểm AB , O thuộc tia đối của AB
Nên : \(OM=OA+MA\)
và : \(OM=OB-MB\)
\(\Rightarrow20M=OA+OB\)
( Vì \(MA=MB\) )
\(\Rightarrow OM=\frac{1}{2}\left(OA+OB\right)\)
TRƯỜNG HỢP 2 :
O thuộc tia đối của 0A ,chứng minh tương tự ta cũng có : \(OM=\frac{1}{2}\left(OA+OB\right)\)
Vậy điểm O không thuộc đoạn thẳng AB thì \(OM=\frac{1}{2}\left(OA+OB\right)\)
Chúc bạn học tốt ( -_- )
lập luân.....để:
OM= OA + BM ( VÌ AM=BM )
+
OM= OB - BM
=> 2.OM= OA + BM + OB - BM
=> OM = ( OA + OB ) : 2