Cho a+b+c=0
C/m \(a^3 + b^3 + c^3 = 3abc\)
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a^3+b^3+c^3-3abc=0
=>(a+b)^3+c^3-3ab(a+b)-3abc=0
=>(a+b+c)(a^2+b^2+2ab-ac-bc+c^2-3ab)=0
=>(a+b+c)(a^2+b^2+c^2-ab-bc-ac)=0
=>a^2+b^2+c^2-ab-bc-ac=0
=>2a^2+2b^2+2c^2-2ab-2bc-2ac=0
=>(a-b)^2+(b-c)^2+(a-c)^2=0
=>a=b=c
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.
\(a^3+b^3+c^3=3abc\)
\(\Rightarrow a^3+b^3+c^3-3abc=0\)
\(\Rightarrow\left(a+b+c\right)\left(a^2+b^2+c^2-ab-ac-bc\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}a+b+c=0\\a^2+b^2+c^2-ab-ac-bc=0\end{matrix}\right.\)
+ \(a^2+b^2+c^2-ab-bc-ac=0\)
\(\Rightarrow2a^2+2b^2+2c^2=2ab+2bc+2ac\)
\(\Rightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\)
\(\Rightarrow\left\{{}\begin{matrix}\left(a-b\right)^2=0\\\left(b-c\right)^2=0\\ \left(c-a\right)^2=0\end{matrix}\right.\)
\(\Rightarrow a=b=c\)
\(a^3+b^3+c^3=3bac\)
=>\(\left(a+b\right)^3-3ab\left(a+b\right)+c^3-3abc=0\)
=>\(\left[\left(a+b\right)^3+c^3\right]-3ba\left(a+b+c\right)=0\)
=>\(\left(a+b+c\right)\left[\left(a+b\right)^2-c\left(a+b\right)+c^2\right]-3ab\left(a+b+c\right)=0\)
=>\(\left(a+b+c\right)\left(a^2+2ab+b^2-ac-bc+c^2-3ab\right)=0\)
=>\(a^2+b^2+c^2-ab-ac-bc=0\)
=>\(2a^2+2b^2+2c^2-2ab-2ac-2bc=0\)
=>\(\left(a^2-2ab+b^2\right)+\left(b^2-2bc+c^2\right)+\left(a^2-2ac+c^2\right)=0\)
=>\(\left(a-b\right)^2+\left(b-c\right)^2+\left(a-c\right)^2=0\)
=>a=b=c
\(a^3+b^3+c^3=3abc\\\Rightarrow a^3+b^3+c^3-3abc=0\\\Rightarrow(a+b)^3+c^3-3ab(a+b)-3abc=0\\\Rightarrow (a+b+c)^3-3(a+b)c(a+b+c)-3ab(a+b+c)=0\\\Rightarrow(a+b+c)[(a+b+c)^2-3(a+b)c-3ab]=0\\\Rightarrow(a+b+c)(a^2+b^2+c^2+2ab+2bc+2ca-3ac-3bc-3ab)=0\\\Rightarrow(a+b+c)(a^2+b^2+c^2-ab-bc-ca)=0\\\Rightarrow a^2+b^2+c^2-ab-bc-ca=0(vì.a+b+c\ne0)\\\Rightarrow2a^2+2b^2+2c^2-2ab-2bc-2ca=0\\\Rightarrow(a^2-2ab+b^2)+(b^2-2bc+c^2)+(c^2-2ca+a^2)=0\\\Rightarrow(a-b)^2+(b-c)^2+(c-a)^2=0\)
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:
\(a^3+b^3=\left(a+b\right)^3-3ab\left(a+b\right)\)
\(\left(a+b+c\right)^2=a^2+b^2+c^2+2ab+2bc+2ca\)
$\text{#}Toru$
Ta có:
x^3 + y^3 + z^3 - 3xyz = (x+y)^3 + z^3 - 3x^2y - 3xy^2 - 3xyz
= (x+y)^3 + z^3 - 3xy(x + y + z)
= (x+y+z)^3 - 3(x+y)^2.z - 3(x+y)z^2 - 3xy(x + y + z)
= (x+y+z)^3 - 3(x+y)z(x+ y + z) - 3xy(x + y + z)
=(x+y+z)[(x+y+z)^2 - 3(x+y)z - 3xy]
Với x+y+z = 0 => x^3 + y^3 + z^3 - 3xyz = 0
=> x^3 + y^3 + z^3 = 3xyz (đpcm)
a + b + c = 0 => a + b = -c
a^3 + b^3 + c^3 = ( a+ b)^3 - 3a^2b - 3ab^2 + c^3
= ( a+ b)^3 + 3ab ( a+b) +c^3
Thay a+ b = - c ta có
a^3 + b^3 +c^3 = -c^3 - 3ab.-c + c^3 = 3abc => ĐPCM
`(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`
`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`
Ta có: \(a^3+b^3+c^3=3abc\)
\(\Rightarrow a^3+b^3+c^3-3abc=0\)
\(\Leftrightarrow\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}a+b+c=0\\a^2+b^2+c^2-ab-bc-ca=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}a+b+c=0\\2a^2+2b^2+2c^2-2ab-2bc-2ca=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}a+b+c=0\\\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}a+b+c=0\\a=b=c\end{cases}}\)
bạn thay vào M giải tiếp nha
Ta có: \(a^3+b^3+c^3=3abc\)
\(\Leftrightarrow\left(a^3+b^3\right)+c^3-3abc=0\)
\(\Leftrightarrow\left(a+b\right)^3-3ab\left(a+b\right)+c^3-3abc=0\)
\(\Leftrightarrow\left[\left(a+b\right)^3+c^3\right]-\left[3ab\left(a+b\right)+3abc\right]=0\)
\(\Leftrightarrow\left(a+b+c\right)\left[\left(a+b\right)^2-\left(a+b\right)c+c^2\right]-3ab\left(a+b+c\right)=0\)
\(\Leftrightarrow\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)=0\)
Nếu \(a^2+b^2+c^2-ab-bc-ca\)
\(=\frac{1}{2}\left[\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\right]\ge0\left(\forall a,b,c\right)\)
Dấu "=" xảy ra khi: a = b = c
Khi đó: \(M=\left(1+\frac{a}{b}\right)\left(1+\frac{b}{c}\right)\left(1+\frac{c}{a}\right)=\left(1+1\right)^3=8\)
Nếu \(a+b+c=0\Rightarrow\hept{\begin{cases}a+b=-c\\b+c=-a\\c+a=-b\end{cases}}\)
\(\Rightarrow M=\frac{\left(a+b\right)\left(b+c\right)\left(c+a\right)}{abc}=\frac{-abc}{abc}=-1\)
1 ) Ta có :
\(a+b-c=0\Leftrightarrow a+b=c\Leftrightarrow\left(a+b\right)^3=c^3\)
\(\Rightarrow a^3+b^3-c^3=a^3+b^3-\left(a+b\right)^3\)
\(\Rightarrow a^3+b^3-c^3=a^3+b^3-3a^2b-3b^2a-b^3\)
\(\Rightarrow a^3+b^3-c^3=-3a^2b-3b^2a\)
\(\Rightarrow a^3+b^3-c^3=-3ab\left(a+b\right)\)
\(\Rightarrow a^3+b^3-c^3=-3abc\left(đpcm\right)\)
2 ) Ta có :
\(a-b+c=0\Leftrightarrow c=b-a\Leftrightarrow c^3=\left(b-a\right)^3\)
\(\Rightarrow a^3-b^3+c^3=a^3-b^3+\left(b-a\right)^3\)
\(\Rightarrow a^3-b^3+c^3=a^3-b^3+b^3-3a^2b+3b^2a-a^3\)
\(\Rightarrow a^3-b^3+c^3=-3a^2b+3b^2a\)
\(\Rightarrow a^3-b^3+c^3=-3ab\left(a-b\right)\)
\(\Rightarrow a^3-b^3+c^3=3ab\left(b-a\right)\)
\(\Rightarrow a^3-b^3+c^3=3abc\left(đpcm\right)\)
1 ) Bổ sung dấu \(\Rightarrow\) thứ 2 :
\(\Rightarrow...=a^3+b^3-a^3-3a^2b-3b^2a-b^3\)
hk rồi mà chẳng hỉu cái quái gì cả hế hế