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Hồ Bảo Ngọc
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Nguyễn Thái Anh
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Vo Tuan Viet
30 tháng 8 2016 lúc 20:15

Bằng nhau

Đỗ Phúc Thiên
30 tháng 8 2016 lúc 21:59

a=b=c=1 suy ra Tam giác ABC là tam giác đều vì có độ dài 3 canh = nhau .

liên hoàng
30 tháng 8 2016 lúc 23:12

ta áp dụng (a+b+c)(\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\)) >=9 

dễ chứng minh bdt phụ này 

rùi từ đây suy ra 3(a-b)(b-c)(c-a) = 0 => a=b=c (1)

mà lên bđt phụ trên thì xảy ra khi a=b=c (1)

từ (1) , (2) , ta suy ra a=b=c hay đpcm 

vì k chặt chẽ lắm nên thông cảm

Tiến Phạm
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anh tuấn
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huongkarry
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Lê Anh Tú
7 tháng 7 2017 lúc 8:18

thực hiện trừ 2 vế ta (vế trái cho vế phải) ta được

(a+b+c).(a^2+b^2+c^2 -ab-bc-ca)=0

nên hoặc a+b+c=0 hoặc nhân tử còn lại bằng 0

mà a,b,c là 3 cạnh 1 tam giác nên a+b+c>0

vậy a^2+b^2+c^2 -ab-bc-bc-ca=0

đặt đa thức đó bằng A

A=0 nên 2xA=0

phân tích thành hằng đẳng thức ta có (a-b)2+(b-c)2+(c-a)2=0

nên a=b=c vậy là tam giác đều 

Akai Haruma
21 tháng 10 2024 lúc 23:02

Lời giải:

$a^3+b^3+c^3=3abc$

$\Leftrightarrow (a+b)^3-3ab(a+b)+c^3-3abc=0$

$\Leftrightarrow (a+b)^3+c^3-3ab(a+b+c)=0$

$\Leftrightarrow (a+b+c)[(a+b)^2-c(a+b)+c^2]-3ab(a+b+c)=0$

$\Leftrightarrow (a+b+c)(a^2+b^2+c^2-ab-bc-ac)=0$

Hiển nhiên $a+b+c>0$ với mọi $a,b,c$ là độ dài 3 cạnh tam giác.

$\Rightarrow a^2+b^2+c^2-ab-bc-ac=0$

$\Leftrightarrow 2a^2+2b^2+2c^2-2ab-2bc-2ac=0$

$\Leftrightarrow (a-b)^2+(b-c)^2+(c-a)^2=0$

Do mỗi số $(a-b)^2; (b-c)^2; (c-a)^2\geq 0$ với mọi $a,b,c>0$.

$\Rightarrow$ để tổng của chúng bằng $0$ thì:

$(a-b)^2=(b-c)^2=(c-a)^2=0$

$\Rightarrow a=b=c$

$\Rightarrow ABC$ là tam giác đều.

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

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Kimian Hajan Ruventaren
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Hồng Phúc
26 tháng 1 2021 lúc 15:23

Ta có \(S=\dfrac{abc}{4R}=pr=\sqrt{p\left(p-a\right)\left(p-b\right)\left(p-c\right)}\)

\(\Rightarrow S^2=\dfrac{abcpr}{4R}=p\left(p-a\right)\left(p-b\right)\left(p-c\right)\)

\(\Rightarrow\dfrac{2r}{R}=\dfrac{\left(a+b-c\right)\left(b+c-a\right)\left(c+a-b\right)}{abc}\)

Theo giả thiết \(\dfrac{a^3+b^3+c^3}{abc}+\dfrac{2r}{R}=4\)

\(\Leftrightarrow\dfrac{a^3+b^3+c^3}{abc}+\dfrac{\left(a+b-c\right)\left(b+c-a\right)\left(c+a-b\right)}{abc}=4\)

\(\Leftrightarrow a^3+b^3+c^3+\left(a+b-c\right)\left(b+c-a\right)\left(c+a-b\right)=4abc\)

\(\Leftrightarrow a^2b+ab^2+b^2c+bc^2+c^2a+ca^2=6abc\left(1\right)\)

Áp dụng BĐT AM-GM:

\(a^2b+ab^2+b^2c+bc^2+c^2a+ca^2\ge6abc\)

\(\Rightarrow\left(1\right)\) đúng

Đẳng thức xảy ra khi \(a=b=c\)

\(\Leftrightarrow\Delta ABC\) đều

Duyên Lương
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Siêu Nhân Lê
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