\(a^3+b^3+c^3=3bac\)

=>\(\left(a+b\right)^3-3ab\left(a+b\right)+c^3-3abc=0\)

=>\(\left[\left(a+b\right)^3+c^3\right]-3ba\left(a+b+c\right)=0\)

=>\(\left(a+b+c\right)\left[\left(a+b\right)^2-c\left(a+b\right)+c^2\right]-3ab\left(a+b+c\right)=0\)

=>\(\left(a+b+c\right)\left(a^2+2ab+b^2-ac-bc+c^2-3ab\right)=0\)

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

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

=>\(\left(a^2-2ab+b^2\right)+\left(b^2-2bc+c^2\right)+\left(a^2-2ac+c^2\right)=0\)

=>\(\left(a-b\right)^2+\left(b-c\right)^2+\left(a-c\right)^2=0\)

=>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:\(a^3+b^3+c^3=3abc\Rightarrow a^3+b^3+c^3-3abc=0\)

\(\Rightarrow a^3+3a^2b+3ab^2+b^3+c^3-3a^2b-3ab^2-3abc=0\)

\(\Rightarrow\left(a+b\right)^3+c^3-3ab\left(a+b+c\right)=0\)

\(\Rightarrow\left(a+b+c\right)\left[\left(a+b\right)^2-\left(a+b\right).c+c^2\right]-3ab\left(a+b+c\right)=0\)

\(\Rightarrow\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)=0\)

\(\Rightarrow\orbr{\begin{cases}a+b+c=0\\a^2+b^2+c^2-ab-bc-ca=0\end{cases}}\)\(\Rightarrow\orbr{\begin{cases}a+b+c=0\left(loai\right)\\a=b=c\end{cases}}\)

\(\Rightarrow P=2007.2007.2007=2007^3\)

\(a^3+b^3+c^3=3abc\\ \Rightarrow a^3+b^3+c^3-3abc=0\\ \Rightarrow\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ac\right)=0\\ \Rightarrow\left[{}\begin{matrix}a+b+c=0\\a^2+b^2+c^2-ab-bc-ac=0\end{matrix}\right.\)

\(\Rightarrow a^2+b^2+c^2=ab+bc+ac\left(a+b+c\ne0\right)\\ \Rightarrow2a^2+2b^2+2c^2=2ab+2bc+2ac\\ \Rightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(a-c\right)^2=0\\ \Rightarrow a=b=c\\ \Rightarrow B=\dfrac{2}{a}.\dfrac{2}{b}.\dfrac{2}{c}=\dfrac{8}{abc}\)

\(a^3+b^3+c^3=3abc\)

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

\(\Leftrightarrow\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)=0\)(tự nhân lại rồi phân tích)

\(\Leftrightarrow\left[{}\begin{matrix}a+b+c=0\\a^2+b^2+c^2-ab-bc-ca=0\end{matrix}\right.\)

+)Xét a+b+c=0\(\Rightarrow P=\dfrac{b+a}{b}\cdot\dfrac{c+b}{c}\cdot\dfrac{a+c}{a}=\dfrac{-c}{b}\cdot\dfrac{-a}{c}\cdot\dfrac{-b}{a}=-1\)

+Xét \(a^2+b^2+c^2-ab-bc-ca=0\)

\(\Leftrightarrow\dfrac{1}{2}\left[\left(a^2-2ab+b^2\right)+\left(b^2-2bc+c^2\right)+\left(c^2-2ca+a^2\right)\right]=0\)

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

\(\Leftrightarrow a=b=c\)

\(\Rightarrow P=2\cdot2\cdot2=8\)

Giải:

Từ \(a^3+b^3+c^3=3abc\Leftrightarrow\)\(\left[{}\begin{matrix}a+b+c=0\\a=b=c\end{matrix}\right.\)

Ta xét các trường hợp:

Trường hợp \(1\): Nếu \(a+b+c=0\) thì:

\(\Rightarrow\left\{{}\begin{matrix}a+b=-c\\b+c=-a\\a+c=-b\end{matrix}\right.\)

Thay vào \(P\) ta có:

\(P=\left(1+\dfrac{a}{b}\right)\left(1+\dfrac{b}{c}\right)\left(1+\dfrac{c}{a}\right)\)

\(=\left(\dfrac{a+b}{b}\right)\left(\dfrac{b+c}{c}\right)\left(\dfrac{a+c}{c}\right)\)

\(=\dfrac{-c}{b}.\dfrac{-a}{c}.\dfrac{-b}{a}=\dfrac{\cdot\left(-c\right).\left(-a\right).\left(-b\right)}{b.c.a}=-1\)

Trường hợp \(2\): Nếu \(a=b=c\) thì:

\(P=\left(1+\dfrac{a}{b}\right)\left(1+\dfrac{b}{c}\right)\left(1+\dfrac{c}{a}\right)\)

\(=\left(1+\dfrac{a}{a}\right)\left(1+\dfrac{a}{a}\right)\left(1+\dfrac{a}{a}\right)\)

\(=\left(1+1\right)\left(1+1\right)\left(1+1\right)\)

\(=2.2.2=8\)

Vậy \(P=-1\) hoặc \(P=8\)

ta có : a³+b³+c³-3abc=0

\(\Rightarrow\)(a+b)³+c³-3abc-3a²b-3ab²=0

\(\Rightarrow\)(a+b+c)(a²+b²+c²+2ab-ac-bc)-3ab(a+b+c)=0

\(\Rightarrow\)(a+b+c)(a²+b²+c²-ab-ac-bc)=0

\(\Rightarrow\)\(\left[{}\begin{matrix}a+b+c=0\\a^2+b^2+c^2-ab-bc-ac=0\end{matrix}\right.\)

\(\Rightarrow\)\(\left[{}\begin{matrix}\left\{{}\begin{matrix}a=-\left(b+c\right)\\b=-\left(a+c\right)\\c=-\left(a+b\right)\end{matrix}\right.\\\left(a+b+c\right)^2+a^2+b^2+c^2=0\Leftrightarrow a=b=c=0\left(bỏ\right)\end{matrix}\right.\)ta có P=(1+\(\dfrac{a}{b}\))(1+\(\dfrac{b}{c}\))(1+\(\dfrac{c}{a}\))

\(\Leftrightarrow\)p=\(\left(\dfrac{b+a}{b}\right)\left(\dfrac{c+b}{c}\right)\left(\dfrac{a+c}{a}\right)\)

\(\Leftrightarrow P=\left(\dfrac{-c}{b}\right)\left(\dfrac{-a}{c}\right)\left(\dfrac{-b}{a}\right)\)

\(\Leftrightarrow\)P=-1

\(\left\{{}\begin{matrix}a^3+b^3+c^3=3abc\\a+b+c\ne0\end{matrix}\right.\) \(\Leftrightarrow a^2+b^2+c^2-\left(ab+bc+ca\right)=0\)

\(P=\dfrac{\left(a^2+b^2+c^2\right)^3}{\left(a+b+c\right)^6}=\left[\dfrac{a^2+b^2+c^2}{\left(a+b+c\right)^2}\right]^3=\left[\dfrac{\left(ab+bc+ca\right)}{3\left(ab+bc+ca\right)}\right]^3=\dfrac{1}{27}\)

Ta có : \(a^3+b^3+c^3-3abc=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-c\right)\)

Do : \(a^3+b^3+c^3=3abc\) và \(a+b+c\ne0\) nên \(a^2+b^2+c^2-ab-bc-ac=0\)

Dễ dàng suy ra \(a=b=c\).Vậy \(N=\frac{3a^2}{\left(3a\right)^2}=\frac{1}{3}.\)

Mẫu của N phải là (a+b+c)^2013 chứ bạn

Đk để phân số tồn tại là : a+b+c khác 0

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

<=> a^3+b^3+c^3-3abc = 0

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

<=> a^2+b^2+c^2-ab-bc-ca = 0 ( vì a+b+c khác 0 )

<=> 2a^2+2b^2+2c^2-2ab-2bc-2ca = 0

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

<=> (a-b)^2+(b-c)^2+(c-a)^2 = 0

<=> a-b=0 ; b-c=0 ; c-a=0

<=> a=b=c

Khi đó : N = 3a^2013/(3a)^2013 = 3/3^2013 = 1/3^2012

Tk mk nha