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Sửa đề: \(AB^2+AC^2=2AM^2+\frac{BC^2}{2}\)
ΔAHC vuông tại H
=>\(AC^2=AH^2+HC^2\)
ΔAHB vuông tại H
=>\(AB^2=AH^2+HB^2\)
\(AB^2+AC^2=AH^2+HB^2+AH^2+HC^2\)
\(=2\cdot AH^2+HB^2+HC^2\)
\(=2HA^2+\left(HM+MB\right)^2+\left(MC-MH\right)^2\)
\(=2HA^2+\left(HM+MB\right)^2+\left(MB-MH\right)^2\)
\(=2HA^2+HM^2+MB^2+2\cdot HM\cdot MB+HM^2+MB^2-2\cdot HM\cdot MB\)
\(=2HA^2+2\cdot HM^2+2\cdot MB^2=2\cdot\left(HA^2+HM^2\right)+2\cdot MB^2\)
\(=2\cdot AM^2+2\cdot\left(\frac{BC}{2}\right)^2=2\cdot AM^2+2\cdot\frac{BC^2}{4}=2\cdot AM^2+\frac{BC^2}{2}\)
Sửa đề: \(AB^2+AC^2=2AM^2+\frac{BC^2}{2}\)
ΔAHC vuông tại H
=>\(AC^2=AH^2+HC^2\)
ΔAHB vuông tại H
=>\(AB^2=AH^2+HB^2\)
\(AB^2+AC^2=AH^2+HB^2+AH^2+HC^2\)
\(=2\cdot AH^2+HB^2+HC^2\)
\(=2HA^2+\left(HM+MB\right)^2+\left(MC-MH\right)^2\)
\(=2HA^2+\left(HM+MB\right)^2+\left(MB-MH\right)^2\)
\(=2HA^2+HM^2+MB^2+2\cdot HM\cdot MB+HM^2+MB^2-2\cdot HM\cdot MB\)
\(=2HA^2+2\cdot HM^2+2\cdot MB^2=2\cdot\left(HA^2+HM^2\right)+2\cdot MB^2\)
\(=2\cdot AM^2+2\cdot\left(\frac{BC}{2}\right)^2=2\cdot AM^2+2\cdot\frac{BC^2}{4}=2\cdot AM^2+\frac{BC^2}{2}\)
a: \(AB^2-BH^2=AH^2\)
\(AC^2-CH^2=AH^2\)
Do đó: \(AB^2-BH^2=AC^2-CH^2\)
=>\(AB^2+CH^2=AC^2+BH^2\)
b: \(AC^2-AB^2=AH^2+HC^2-AH^2-HB^2\)
\(=HC^2-HB^2=2\cdot BC\cdot HM\)
Nếu đến tối nay mà còn bí thì hú mình. Mình không hứa sẽ làm được bài này nhưng hứa sẽ suy nghĩ cùng b :p
ui bài này dễ thế mà cậu k biết làm à
Trương Ngọc Uyển Nhi. Chính vì k bít nên ms pải đăng lên đây đó cậu bít thì vô làm đi..
ko giải được thì bớt xàm nhé bạn trẻ
Hoàng Phúc nói mk sao b??
Không phải Hoàng Phúc nói b đâu. Nói bạn bên dưới đó. Mà thấy đề ngăn ngắn chắc là không khó. Bạn thử tự làm xem sao :)
Khó dẽ là hàm phụ thuộc rất nhiều biến
có thể khó với A nhưng lại cực kỳ khó B và cực dẽ với B
Lời giải
đặt
A=duong thi minh
B=Ailibaba
C=Trương Ngọc Uyen nhi
Từ đầu bài Ta có phương trình cân bằng
\(AB^2=AC^2+2BC.MH\)
Thiếu Biến => Bổ thêm
Với ngonhuminh trung lập => ngonhuminh <=> trung tuyến
Vậy ngonhuminh =M trong phương trình trên
tất nhiên Hoang phúc là H
Khá đấy ngonhuminh à
Khó dẽ là hàm phụ thuộc rất nhiều biến
có thể khó với A nhưng lại cực kỳ khó B và cực dẽ với B
Lời giải
đặt
A=duong thi minh
B=Ailibaba
C=Trương Ngọc Uyen nhi
Từ đầu bài Ta có phương trình cân bằng
$AB^2=AC^2+2BC.MH$AB2=AC2+2BC.MH
Thiếu Biến => Bổ thêm
Với ngonhuminh trung lập => ngonhuminh <=> trung tuyến
Vậy ngonhuminh =M trong phương trình trên
tất nhiên Hoang phúc là H
k mình nha
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tìm chữ cái khác trong những chữ này và nêu đó là chữ gì?
Bạn tự vẽ hình nha.
b) AB2 - AC2 = (BH2 + AH2) - (CH2 + AH2) = BH2 - CH2 = (BH+CH)(BH-CH) = BC.[(BM+MH)-(CM-MH)]=BC.2MH (do BM=CM).
Còn phần a bạn vào link này nha: https://diendan.hocmai.vn/threads/toan-9-chung-minh-cong-thuc-tinh-duong-trung-tuyen-trong-tam-giac.355106/.
Chúc bạn học tốt.