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

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

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

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

Áp dụng tính chất của dãy tỉ số bằng nhau , ta có

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

\(\Rightarrow\hept{\begin{cases}a+b=2c\\b+c=2a\\a+c=2b\end{cases}}\)

Vậy \(P=\left(a+b\right)\left(b+c\right)\left(c+a\right)=2c.2a.2b=8abc\)

mà \(\left(a+b\right)\left(b+c\right)\left(c+a\right)=abc\Rightarrow8abc=abc\Rightarrow abc=0\Rightarrow P=0\)

Ta có: a³+b³+c³=3abc <=> a³+b³+c³-3abc=0

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

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

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

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

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

Mà a+b+c khác 0

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

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

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

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

<=>\(\hept{\begin{cases}\left(a-b\right)^2=0\\\left(b-c\right)^2=0\\\left(c-a\right)^2=0\end{cases}\Leftrightarrow\hept{\begin{cases}a-b=0\\b-c=0\\c-a=0\end{cases}\Leftrightarrow\hept{\begin{cases}a=b\\b=c\\c=a\end{cases}\Leftrightarrow}}a=b=c}\)

=>\(N=\frac{a^2+b^2+c^2}{\left(a+b+c\right)^2}=\frac{3a^2}{\left(3a\right)^2}=\frac{3a^2}{9a^2}=\frac{1}{3}\)

- Ta có : \(a^3+b^3+c^3=3abc\)

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

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

Mà \(a+b+c\ne0\)

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

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

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

=> \(a-b=b-c=c-a=0\)

=> \(a=b=c\)

- Thay a = b = c vào biểu thức N ta được :

\(N=\frac{a^2+a^2+a^2}{\left(a+a+a\right)^2}=\frac{3a^2}{9a^2}=\frac{1}{3}\)

Vậy giá trị của N = \(\frac{1}{3}\) khi \(a^3+b^3+c^3=3abc\) và \(a+b+c\ne0\)

GIAI NHANH LEN  MINH CAN GAP 

Ta có:\(m=\dfrac{b+c}{a}+\dfrac{c+a}{b}+\dfrac{a+b}{c}\)

\(m=\left(\dfrac{b+c}{a}+1\right)+\left(\dfrac{c+a}{b}+1\right)+\left(\dfrac{a+b}{c}+1\right)-3\)

\(m=\dfrac{a+b+c}{a}+\dfrac{a+b+c}{b}+\dfrac{a+b+c}{c}-3\)

\(m=\left(a+b+c\right)\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)-3\)

\(m=0-3=-3\)

bai nay mk lm dc ...........nhug hoi dai

Từ a + b +c =0 <=>
{a+b = -c = => (a+b)² = c²
{a+c = -b= = > (a+c)² = b²
{b+c = -a = => (b+c)² = a²
<=>
{a² + b² + 2a.b = c²
{a² + c² + 2a.c = b²
{b² + c² + 2b.c = a²
<=>

{a² + b² - c² = -2a.b
{a² + c² - b² = -2a.c
{b² + c² - a² = -2b.c

Lúc đó P viết lại thành

P = 1/(a² + b² - c²) + 1/(a² + c² - b²) + 1/(b² + c² - a² )

P = -(1/ab + 1/ac + 1/bc ) / 2

P = -[( 1/ab + 1/ac ) + ( 1/ab +1/bc) + (1/ac +1/bc)]

P = -[(b+c)/abc + (a+c)/abc + (a+b)/abc]

P = -2(a+b+c)/abc =0