Do \(a,b,c\) là các số dương suy ra:

\(a>0;b>0;c>0\)

Suy ra: \(a+b+c>0\)

Ta có: \(a^3+b^3+c^3=3abc\)

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

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

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

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

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

Do \(a+b+c>0\)

Suy ra: \(a^2+b^2+c^2-ab-bc-ca=0\)

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

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

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

Suy ra: \(a-b=0;b-c=0\) và \(c-a=0\)

Suy ra: \(a=b=c\)

Suy ra: \(\frac{a}{b}=\frac{b}{c}=\frac{c}{a}=1\)

Ta có: \(\left(\frac{a}{b}-1\right)+\left(\frac{b}{c}-1\right)+\left(\frac{c}{a}-1\right)=\left(1-1\right)+\left(1-1\right)+\left(1-1\right)=0\)

Vậy ...

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 \(a^2+b^2+c^2-ab-bc-ca=0\)

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

\(\Rightarrow a=b=c\)

Khi đó \(A=2^3=8\)

Nếu \(a+b+c=0\Rightarrow a+b=-c;b+c=-a;c+a=-b\)

Thay vào ta được:

\(A=\frac{a+b}{b}\cdot\frac{b+c}{c}\cdot\frac{c+a}{a}=\frac{-abc}{abc}=-1\)

Vậy A = 8 hoặc A = -1

Theo tính chất dãy tỉ số bằng nhau ta có : a+b-c/c = b+c-a/a = c+a-b/b = a+b-c+b+c-a+c+a-b/a+b+c = a+b+c/a+b+c = 1

Ta có : a+b-c/c=1  => a+b-c=c  => a+b+c=3c   (1)

Ta có : b+c-a/a=1  => b+c-a=a  => a+b+c=3a   (2)

Ta có : c+a-b/b=1  => c+a-b=b  => a+b+c=3b   (3)

Từ (1);(2);(3)   => 3c=3a=3b  => a=b=c  => b/a=1 ; a/c=1 ; c/b=1

=> B= (1+b/a)(1+a/c)(1+c/b)  = (1+1)(1+1)(1+1) = 2.2.2 = 8

=8

8 8 cái địt mẹ mày

Ta có:

(a+b-c)/c=(b+c-a)/a=(c+a-b)/b=(a+b-c+b+c-a+c+a-b)/(c+a+b)=0/(c+a+b)=0

=> a+b-c=0 =>a+b=c

b+c-a=0 =>b+c=a

c+a-b=0 =>c+a=b

=>B=(a+b)/a.(c+a)/c.(b+c)/b

      =c/a.b/c.a/b=1

TK!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

Ta có:

(a+b-c)/c=(b+c-a)/a=(c+a-b)/b=(a+b-c+b+c-a+c+a-b)/(c+a+b)=0/(c+a+b)=0

=> a+b-c=0 =>a+b=c

b+c-a=0 =>b+c=a

c+a-b=0 =>c+a=b

=>B=(a+b)/a.(c+a)/c.(b+c)/b

      =c/a.b/c.a/b=1

Ta có : \(\left(1+\frac{b}{a}\right)\cdot\left(1+\frac{a}{c}\right)\cdot\left(1+\frac{c}{b}\right)\)=\(\frac{a+b}{a}\cdot\frac{a+c}{c}\cdot\frac{b+c}{b}\)(*)

\(\frac{a+b-c}{c}=\frac{b+c-a}{a}=\frac{c+a-b}{b}\)

=> \(\frac{a+b}{c}-1=\frac{b+c}{a}-1=\frac{c+a}{b}-1\)

=>\(\frac{a+b}{c}=\frac{b+c}{a}=\frac{c+a}{b}\)

ADTC dãy tỉ số = nhau ta có:

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

Suy ra : \(\frac{a+b}{c}\)=2  => a+b=2c   (1)

\(\frac{b+c}{a}\)=2  => b+c=2a   (2)

\(\frac{c+a}{b}\)=2  => c+a=2b   (3)

Thay (1);(2);(3) vào (*) ta có :

\(\frac{2c}{a}\cdot\frac{2b}{c}\cdot\frac{2a}{b}\)=\(\frac{\left(2\cdot2\cdot2\right)\cdot\left(a\cdot b\cdot c\right)}{a\cdot b\cdot c}\)=8