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Trần Hoàng Minh
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Nguyễn Lê Phước Thịnh
9 tháng 10 2021 lúc 21:37

b: Xét ΔABH vuông tại H và ΔACK vuông tại K có

AB=AC

\(\widehat{A}\) chung

Do đó: ΔABH=ΔACK

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Tấn Sang Nguyễn
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Nguyễn Lê Phước Thịnh
12 tháng 5 2023 lúc 9:02

ΔAHB vuông tại H có HE vuông góc AB

nên AH^2=AE*AB

ΔAHC vuông tại H có HD vuông góc AC

nên AH^2=AD*AC

=>AE*AB=AD*AC

=>AE/AC=AD/AB

=>ΔAED đồng dạng với ΔACB

=>góc AED=góc ACB

=>góc BED+góc BCD=180 độ

=>góc CDE+góc B=180 độ

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dsfdsf
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Trịnh Việt Dũng
15 tháng 6 2022 lúc 20:31

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nguyen yen nhi
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jdbcjkervkver
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Mori Ran
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Nguyễn Lê Phước Thịnh
19 tháng 11 2022 lúc 23:19

loading...

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Lê Văn Tâm
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Minh Triều
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Trần Đức Thắng
10 tháng 3 2016 lúc 20:18

BAC = 90 độ 

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1st_Parkour
10 tháng 3 2016 lúc 20:19

góc BAC=90 độ. 

nha

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Nguyễn Đại Dương
10 tháng 3 2016 lúc 20:22

BAC = 90o

Đáp án chính xác

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Nuyen Thanh Dang
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Phước Nguyễn
10 tháng 7 2016 lúc 22:26

  Đã xảy ra lỗi rồi. Bạn thông cảm vì sai sót này.

  Ta có:  

Áp dụng hệ quả của bất đẳng thức Cauchy cho ba số không âm 

   trong đó với     , ta có:

  

Tương tự, ta có:

       

Cộng ba bất đẳng thức     và   , ta được:

  

Khi đó, ta chỉ cần chứng minh

  

Thật vậy, bất đẳng thức cần chứng minh được quy về dạng sau:    (bất đẳng thức Cauchy cho ba số   )

Hay       

Mà    đã được chứng minh ở câu    nên    luôn đúng với mọi  

Dấu    xảy ra    

Vậy,       

 
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Phương Thảo
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Trịnh Việt Dũng
15 tháng 6 2022 lúc 20:29

chịu hoi =))))))

 

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Trịnh Việt Dũng
15 tháng 6 2022 lúc 20:29

em mới học lớp 7 hà

năm nay lên lớp 8 =)))))

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Nguyễn Thảo My
14 tháng 1 2023 lúc 21:25

1)Ta có: \(S_{ABC}=\dfrac{1}{2}AB.AC.\sin A\)

\(\Leftrightarrow8=\dfrac{1}{2}\times4\times5\times sinA\)

\(\Leftrightarrow\sin A=0,8\)

Lại có: \(\left(\sin A\right)^2+\left(\cos A\right)^2=1\Leftrightarrow\cos A=0,6.\)

Áp dụng định lí hàm số cosin:

\(BC^2=AB^2+AC^2-2AB\times AC\times\cos A\)

\(\Leftrightarrow BC^2=4^2+5^2-2\times4\times5\times0,6=17\)

\(\Leftrightarrow BC=\sqrt{17}.\)

2) Trong \(\Delta ABC\) có: \(g\text{ó}cA+g\text{óc}B+g\text{óc}C=180^o\)

=> BAC=75o.

Áp dụng định lí hàm số sin:

\(\dfrac{AB}{\sin C}=\dfrac{BC}{\sin A}\Leftrightarrow\dfrac{3}{\sin45^o}=\dfrac{BC}{\sin75^o}\)

\(\Leftrightarrow BC=\dfrac{3+3\sqrt{3}}{2}\).

 

 

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