Ta có: \(a^3+b^3+c^3=3abc\)

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

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

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

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

mà \(a+b+c\ne0\)

nên \(a^2+b^2+c^2-ab-ac-bc=0\)

\(\Leftrightarrow2a^2+2b^2+2c^2-2ab-2ac-2bc=0\)

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

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

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

Ta có: \(M=\dfrac{a^{2020}+b^{2020}+c^{2020}}{\left(a+b+c\right)^{2020}}\)

\(=\dfrac{a^{2020}+a^{2020}+a^{2020}}{\left(a+a+a\right)^{2020}}=\dfrac{3\cdot a^{2020}}{9\cdot a^{2020}}=\dfrac{1}{3}\)

Đoạn cuối em bị nhầm rồi kìa. \(\frac{a^{2020}+b^{2020}+c^{2020}}{(a+b+c)^{2020}}=\frac{3a^{2020}}{(3a)^{2020}}=\frac{3}{3^{2020}}=\frac{1}{3^{2019}}\)

Ta có : a³ + b³ + c³ = 3abc

=> (a + b)(a² - ab + b²) + c³ - 3abc = 0

=> (a + b)³ - 3ab(a + b) + c³ - 3abc = 0

=> [(a + b)³ + c³] - [(3ab(a + b) + 3abc] = 0

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

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

=> a² + b² + c² - ab- ac - bc = 0

=> 2(a² + b² + c² - ab- ac - bc) = 0

=> 2a² + 2b² + 2c² - 2ab - 2ac - 2bc = 0

=> (a² - 2ab + b²) + (b² - 2bc + c²) + (a² - 2ac + c²) = 0

=> (a - b)² + (b - c)² + (a - c)² = 0

=> \(\hept{\begin{cases}a-b=0\\b-c=0\\c-a=0\end{cases}}\Rightarrow a=b=c\)

Khi đó M = \(\frac{a^{2020}+b^{2020}+c^{2020}}{\left(a+b+c\right)^{2020}}=\frac{3.c^{2020}}{\left(3c\right)^{2020}}+\frac{3c^{2020}}{3^{2020}.c^{2020}}=\frac{1}{3^{2019}}\)

Bạn ơi! ABC khác 0 thì làm sao ạ+b+c=0 được bạn

           Bài làm :

Vì :

 \(a+b+c=0\)

\(\Rightarrow\hept{\begin{cases}a+b=-c\\b+c=-a\\a+c=-b\end{cases}}\)

Ta có :

 \(A=\left(1+\frac{a}{b}\right)\left(1+\frac{b}{c}\right)\left(1+\frac{c}{a}\right)\)

\(\Rightarrow A=\frac{b+a}{b}.\frac{c+b}{c}.\frac{a+c}{a}\)

\(\Rightarrow A=\left(-\frac{c}{b}\right).\left(\frac{-a}{c}\right).\left(\frac{-b}{a}\right)\)

\(\Rightarrow A=-\frac{abc}{abc}\)

\(\Rightarrow A=-1\)

Vậy A=-1

a:b=b:c=c:a=>a/b=b/c=c/a=a+b+c/b+c+a=1

suy ra: a/b=1 suy ra: a=b

b/c=1 =>b=c

suy ra: a=b=c

suy ra: a^2.b^2.c^1930:b^1935=1.1.1:1=1

sao ko aj trả lời za