Có:

\(a^3+b^3+c^3=3abc\\\Leftrightarrow a^3+b^3+c^3-3abc=0\\\Leftrightarrow (a+b)^3+c^3-3ab(a+b)-3abc=0\\\Leftrightarrow (a+b+c)^3-3(a+b)c(a+b+c)-3ab(a+b+c)=0\\\Leftrightarrow (a+b+c)[(a+b+c)^2-3(a+b)c-3ab]=0\\\Leftrightarrow (a+b+c)(a^2+b^2+c^2+2ab+2bc+2ac-3ac-3bc-3ab)=0\\\Leftrightarrow (a+b+c)(a^2+b^2+c^2-ab-bc-ac)=0\\\Leftrightarrow a^2+b^2+c^2-ab-bc-ac=0(vì.a+b+c\ne0)\\\Leftrightarrow 2a^2+2b^2+2c^2-2ab-2bc-2ac=0\\\Leftrightarrow (a^2-2ab+b^2)+(b^2-2bc+c^2)+(a^2-2ac+c^2)=0\\\Leftrightarrow (a-b)^2+(b-c)^2+(a-c)^2=0\)

Ta thấy: \(\left\{{}\begin{matrix}\left(a-b\right)^2\ge0\forall a,b\\\left(b-c\right)^2\ge0\forall b,c\\\left(a-c\right)^2\ge0\forall a,c\end{matrix}\right.\)

\(\Rightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(a-c\right)^2\ge0\forall a,b,c\)

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

nên: \(\left\{{}\begin{matrix}a-b=0\\b-c=0\\a-c=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=b\\b=c\\a=c\end{matrix}\right.\Leftrightarrow a=b=c\)

Thay \(a=b=c\) vào \(A\), ta được:

\(A=\dfrac{\left(2016+\dfrac{a}{a}\right)+\left(2016+\dfrac{b}{b}\right)+\left(2016+\dfrac{c}{c}\right)}{2017^3}\left(a,b,c\ne0\right)\)

\(=\dfrac{2016+1+2016+1+2016+1}{2017^3}\)

\(=\dfrac{2016\cdot3+1\cdot3}{2017^3}\)

\(=\dfrac{3\cdot\left(2016+1\right)}{2017^3}\)

\(=\dfrac{3}{2017^2}\)

Vậy: ...

ĐKXĐ: \(abc\ne0\)

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

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

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

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

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

TH1: \(a+b+c=0\)

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

TH2: \(a=b=c\Rightarrow P=\left(1+1\right)\left(1+1\right)\left(1+1\right)=8\)

tồn tại A => a,b,c khác 0

=> a+b+c=0

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

TH1 : a + b + c ≠ 0

Áp dụng t/c dãy tỉ số bằng nhau ta có

\(\dfrac{a+b}{c}=\dfrac{b+c}{a}=\dfrac{c+a}{b}=\dfrac{a+b+b+c+a+c}{a+b+c}=2\)

\(\Rightarrow\left\{{}\begin{matrix}a+b=2c\\b+c=2a\\a+c=2b\end{matrix}\right.\)

Khi đó \(M=\left(1+\dfrac{a}{b}\right)\left(1+\dfrac{b}{c}\right)\left(1+\dfrac{c}{a}\right)\)

\(=\dfrac{a+b}{b}.\dfrac{b+c}{c}.\dfrac{a+c}{a}=\dfrac{2c}{b}.\dfrac{2a}{c}.\dfrac{2b}{a}=8\)

TH2 : a + b + c = 0

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

Khi đó \(M=\left(1+\dfrac{a}{b}\right)\left(1+\dfrac{b}{c}\right)\left(1+\dfrac{c}{a}\right)\)

\(=\dfrac{a+b}{b}.\dfrac{b+c}{c}.\dfrac{a+c}{a}=\dfrac{-c}{b}.\dfrac{-a}{c}.\dfrac{-b}{a}=-1\)

Xét 2 TH sau:

TH1: a+b+c=0

Khi đó:

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

TH2: a+b+c khác 0

Ta có:

\(\dfrac{a+b}{c}=\dfrac{b+c}{a}=\dfrac{c+a}{b}=\dfrac{2\left(a+b+c\right)}{a+b+c}=2\)

Suy ra: a+b=2c; b+c=2a; c+a=2b

Do đó:

\(M=\left(1+\dfrac{a}{b}\right)\left(1+\dfrac{b}{c}\right)\left(1+\dfrac{c}{a}\right)\\ =\dfrac{a+b}{b}.\dfrac{b+c}{c}.\dfrac{c+a}{a}\\ =\dfrac{2c}{b}.\dfrac{2a}{c}.\dfrac{2b}{a}\\ =8\)

Xét 2 TH sau:

TH1: a+b+c=0

Khi đó:

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

TH2: a+b+c khác 0

Ta có:

\(\dfrac{a+b}{c}=\dfrac{b+c}{a}=\dfrac{c+a}{b}=\dfrac{2\left(a+b+c\right)}{a+b+c}=2\)

Suy ra: a+b=2c; b+c=2a; c+a=2b

Do đó:

\(M=\left(1+\dfrac{a}{b}\right)\left(1+\dfrac{b}{c}\right)\left(1+\dfrac{c}{a}\right)\\ =\dfrac{a+b}{b}.\dfrac{b+c}{c}.\dfrac{c+a}{a}\\ =\dfrac{2c}{b}.\dfrac{2a}{c}.\dfrac{2b}{a}\\ =8\)

Bổ sung cho bạn Lương Thị Quỳnh Trang

Đặt \(\dfrac{a+b}{c}=\dfrac{b+c}{a}=\dfrac{c+a}{b}=k\left(k\in R\right)\)

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

Cộng 3 đẳng thức trên, ta có:

2(a + b + c) = (a + b + c)k

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

\(\Rightarrow\left[{}\begin{matrix}a+b+c=0\\k=2\end{matrix}\right.\)

Với a + b + c = 0 thì giải như bạn ở dưới

Với k = 2 \(\Rightarrow\left\{{}\begin{matrix}a+b=2c\\b+c=2a\\c+a=2b\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a+b+c=3c\\a+b+c=3a\\a+b+c=3b\end{matrix}\right.\)

=> 3a = 3b = 3c (= a + b + c) <=> a = b = c 

\(M=\left(1+\dfrac{a}{b}\right)\left(1+\dfrac{b}{c}\right)\left(1+\dfrac{c}{a}\right)=2.2.2=8\)

Vậy M = 8

Ta có: \(\dfrac{a+b}{c}=\dfrac{b+c}{a}=\dfrac{c+a}{b}\)\(=\dfrac{2\left(a+b+c\right)}{a+b+c}=2\)

=> a+b=2c; b+c=2a; c+a=2b

Thay vào A ta được: A=((a+b)/b)((c+b)/c)((a+c)/a)

=2c/b.2a/c.2b/a=2.2.2=8

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

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