Giải
Xét \(a^{^{ }3}+b^3+c^3=3abc\) => \(a^3+b^3+c^3-3abc=0\)
=> \(\left(a+b\right)^3-3a^2b-3ab^2-3abc+c^3=0\)
=> \(\left(a+b\right)^3+c^3-3ab\left(a+b+c\right)=0\)
=> \(\left(a+b+c\right)\left[\left(a+b\right)^2-\left(a+b\right)c+c^2\right]-3ab\left(a+b+c\right)=0\)
=> \(\left(a+b+c\right)\left[\left(a+b\right)^2-\left(a+b\right)c+c^3-3ab\right]=0\)
=> \(\left(a+b+c\right)\left(a^2+b^2+c^2-ab-ac-bc\right)=0\)
=>\(\frac{1}{2}\left(a+b+c\right)\left[\left(a^2-2ab+b^2\right)+\left(a^2-2ab+c^2\right)+\left(b^2-2ab+c^2\right)\right]=0\)
=> \(\frac{1}{2}\left(a+b+c\right)\left[\left(a-b\right)^2+\left(a-c\right)^2+\left(b-c\right)^2\right]=0\)
Vì a + b + c = 0 nên \(\left(a-b\right)^2+\left(a-c\right)^2+\left(b-c\right)^2=0\)
a - b = 0
=> a - c = 0 => a = b = c
b - c = 0
Vậy \(M=\left(1+\frac{a}{b}\right)\left(1+\frac{b}{c}\right)\left(1+\frac{c}{a}\right)=8\)
Theo mình nghĩ thì đk sai
:(( Chả bt sao nữa
Quên :
Không sai âu bạn nhé ;) Chỉ có cái đoạn a+b+c=0 thừa hoi
Giải :
Ta có :
\(a^3+b^3+c^3=3abc\)
\(\Leftrightarrow a^3+b^3+c^3-3abc=0\)
\(\Leftrightarrow\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)=0\)
\(\Leftrightarrow\left[\begin{matrix}a+b+c=0\\a^2+b^2+c^2-ab-bc-ca=0\end{matrix}\right.\)
\(\left(+\right)a^2+b^2+c^2-ab-bc-ca=0\)
\(\Rightarrow2\left(a^2+b^2+c^2-ab-bc-ca\right)=0\)
\(\Rightarrow\left(a^2-2ab+b^2\right)+\left(b^2-2bc+c^2\right)+\left(c^2-2ac+a^2\right)=0\)
\(\Rightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\)
Mặt khác :\(\left\{\begin{matrix}\left(a-b\right)^2\ge0\\\left(b-c\right)^2\ge0\\\left(c-a\right)^2\ge0\end{matrix}\right.\)\(\forall a;b;c\)
Giải các ra ta được : a=b=c
=> M = 8
(+) a+b+c=0
\(\Rightarrow\left\{\begin{matrix}a+b=-c\\b+c=-a\\c+a=-b\end{matrix}\right.\)
Mà M \(=\frac{\left(a+b\right)\left(b+c\right)\left(c+a\right)}{abc}\)
=> M = - 1
Mình bổ sung 1 cách khác nhé
a + b + c = 0
<=> (a + b + c)3 = 0
<=> a3 + b3 + c3 + 3(a + b)(b + c)(c + a) = 0
<=> a3 + b3 + c3 = -3(a + b)(b + c)(c + a)
Kết hợp với đề bài: a3 + b3 + c3 = 3abc
Suy ra -3(a + b)(b + c)(c + a) = 3abc
<=> (a + b)(b + c)(c + a)/abc = -1
<=> (a + b)/b . (b + c)/c . (c + a)/a = -1
<=> (1 + a/b)(1 + b/c)(1 + c/a) = -1
=> M = -1
