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M
9
VT
18 tháng 8 2018
Ta có \(a^3+b^3+c^3-3abc=\left(a+b\right)^3+c^3-3ab\left(a+b\right)-3abc=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)\)=0
=> a^3+b^3+c^3=3abc(ĐPCM )
^.^
28 tháng 6 2021
`(a+b+c)^2=3(ab+bc+ca)`
`<=>a^2+b^2+c^2+2ab+2bc+2ca=3(ab+bc+ca)`
`<=>a^2+b^2+c^2=ab+bc+ca`
`<=>2a^2+2b^2+2c^2=2ab+2bc+2ca`
`<=>(a-b)^2+(b-c)^2+(c-a)^2=0`
`VT>=0`
Dấu "=" xảy ra khi `a=b=c`
28 tháng 6 2021
`a^3+b^3+c^3=3abc`
`<=>a^3+b^3+c^3-3abc=0`
`<=>(a+b)^3+c^3-3abc-3ab(a+b)=0`
`<=>(a+b)^3+c^3-3ab(a+b+c)=0`
`<=>(a+b+c)(a^2+b^2+c^2-ab-bc-ca)=0`
`**a+b+c=0`
`**a^2+b^2+c^2=ab+bc+ca`
`<=>a=b=c`
CS
1
15 tháng 12 2016
1) Có: \(a+b+c=0\)
\(\Leftrightarrow a+b=-c\)
\(\Leftrightarrow\left(a+b\right)^3=-c^3\)
\(\Leftrightarrow a^3+b^3+3ab\left(a+b\right)=-c^3\)
\(\Leftrightarrow a^3+b^3-3abc=-c^3\)
\(\Leftrightarrow a^3+b^3+c^3=3abc\)
2)Có: \(a+b-c=0\)
\(\Leftrightarrow a+b=c\)
\(\Leftrightarrow\left(a+b\right)^3=c^3\)
\(\Leftrightarrow a^3+b^3+3ab\left(a+b\right)=c^3\)
\(\Leftrightarrow a^3+b^3+3abc=c^3\)
\(\Leftrightarrow a^3+b^3-c^3=-3abc\)
H
28 tháng 10 2018
1. \(a^3+b^3+c^3-3abc\)
\(=\left(a+b\right)^3+c^3-3ab\left(a+b\right)-3abc\)
\(=\left(a+b+c\right)\left[\left(abc\right)^2-\left(a+b\right)c+c^2\right]-3ab\left(a+b+c\right)\)
\(=\left(a+b+c\right)\left(a^2+2ab+b^2+c^2-ac-bc\right)-3ab\left(a+b+c\right)\)
\(\left(a+b+c\right)\left(a^2+b^2+c^2-ac-bc+2ab-3ab\right)\)
\(=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-ac-bc\right)\)
2. \(a+b+c=0\)
\(\Rightarrow\left(a+b+c\right)\left(a^2+b^2+c^2-ab-ac-bc\right)=0\)
\(\Rightarrow a^3+b^3+c^3-3abc=0\)
\(\Rightarrow a^3+b^3+c^3=3abc\)
3.Còn có a + b + c = 0 nữa mà bn.
\(a^3+b^3+c^3=3abc\)
\(\Rightarrow a^3+b^3+c^3-3abc=0\)
\(\Rightarrow\left(a+b+c\right)\left(a^2+b^2+c^2-ab-ac-bc\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}a+b+c=0\\a^2+b^2+c^2-ab-ac-bc=0\end{matrix}\right.\)
+ \(a^2+b^2+c^2-ab-bc-ac=0\)
\(\Rightarrow2a^2+2b^2+2c^2=2ab+2bc+2ac\)
\(\Rightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\)
\(\Rightarrow\left\{{}\begin{matrix}\left(a-b\right)^2=0\\\left(b-c\right)^2=0\\ \left(c-a\right)^2=0\end{matrix}\right.\)
\(\Rightarrow a=b=c\)
TD
2
D
2
11 tháng 12 2023
\(a^3+b^3+c^3=3bac\)
=>\(\left(a+b\right)^3-3ab\left(a+b\right)+c^3-3abc=0\)
=>\(\left[\left(a+b\right)^3+c^3\right]-3ba\left(a+b+c\right)=0\)
=>\(\left(a+b+c\right)\left[\left(a+b\right)^2-c\left(a+b\right)+c^2\right]-3ab\left(a+b+c\right)=0\)
=>\(\left(a+b+c\right)\left(a^2+2ab+b^2-ac-bc+c^2-3ab\right)=0\)
=>\(a^2+b^2+c^2-ab-ac-bc=0\)
=>\(2a^2+2b^2+2c^2-2ab-2ac-2bc=0\)
=>\(\left(a^2-2ab+b^2\right)+\left(b^2-2bc+c^2\right)+\left(a^2-2ac+c^2\right)=0\)
=>\(\left(a-b\right)^2+\left(b-c\right)^2+\left(a-c\right)^2=0\)
=>a=b=c
T
11 tháng 12 2023
\(a^3+b^3+c^3=3abc\\\Rightarrow a^3+b^3+c^3-3abc=0\\\Rightarrow(a+b)^3+c^3-3ab(a+b)-3abc=0\\\Rightarrow (a+b+c)^3-3(a+b)c(a+b+c)-3ab(a+b+c)=0\\\Rightarrow(a+b+c)[(a+b+c)^2-3(a+b)c-3ab]=0\\\Rightarrow(a+b+c)(a^2+b^2+c^2+2ab+2bc+2ca-3ac-3bc-3ab)=0\\\Rightarrow(a+b+c)(a^2+b^2+c^2-ab-bc-ca)=0\\\Rightarrow a^2+b^2+c^2-ab-bc-ca=0(vì.a+b+c\ne0)\\\Rightarrow2a^2+2b^2+2c^2-2ab-2bc-2ca=0\\\Rightarrow(a^2-2ab+b^2)+(b^2-2bc+c^2)+(c^2-2ca+a^2)=0\\\Rightarrow(a-b)^2+(b-c)^2+(c-a)^2=0\)
Ta thấy: \(\left\{{}\begin{matrix}\left(a-b\right)^2\ge0\forall a,b\\\left(b-c\right)^2\ge0\forall b,c\\\left(c-a\right)^2\ge0\forall a,c\end{matrix}\right.\)
\(\Rightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\ge0\forall a,b,c\)
Mà: \(\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\)
nên: \(\left\{{}\begin{matrix}a-b=0\\b-c=0\\c-a=0\end{matrix}\right.\Rightarrow a=b=c\)
Vậy: ...
\(---\)
Các HĐT được sử dụng trong bài:
\(a^3+b^3=\left(a+b\right)^3-3ab\left(a+b\right)\)
\(\left(a+b+c\right)^2=a^2+b^2+c^2+2ab+2bc+2ca\)
$\text{#}Toru$
D
2
BT
27 tháng 12 2020
a +b +c=0
⇔\(\left(a+b+c\right)^3\)
⇔\(a^3+b^3+c^3+3a^2b+3ab^2+3b^2c+3bc^2+3a^2c+3ac^2+6abc=0\)
⇔\(a^3+b^3+c^3+\left(3a^2b+3ab^2+3abc\right)+\left(3b^2c+3bc^2+3abc\right)+\left(3a^2c+3ac^2+3abc\right)-3abc=0\)
⇔ \(a^3+b^3+c^3+3ab\left(a+b+c\right)+3bc\left(a+b+c\right)+3ac\left(a+b+c\right)=3abc\)
Vì a+b+c= 0
⇒\(a^3+b^3+c^3=3abc\)
Chúc bạn học tốt!
vì a +b+ c = 0
<=> a+b = -c
<=> \(\left(a+b\right)^3=-c^3\)
<=> \(a^3+3a^2b+3ab^2+b^3=-c^3\)
<=> \(a^3+b^3+c^3=-3ab\left(a+b\right)\)
\(a^3+b^3+c^3=3abc\)