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

Do đó \(M=\frac{-c}{c}+\frac{-a}{a}+\frac{-b}{b}=-1+\left(-1\right)+\left(-1\right)=-3\)

Ta có :\(a+b+c=0=>a+b=-c;b+c=-a;a+c=-b\)

Do đó : \(M=-\frac{c}{c}+-\frac{b}{b}+-\frac{c}{c}=\left(-1\right)+\left(-1\right)+\left(-1\right)=-3\)

\(\frac{a}{b}=\frac{b}{c}=\frac{c}{d}=\frac{d}{a}=\frac{a+b+c+d}{a+b+c+d}=1\left(\text{ vì a+b+c+d khác 0}\right)\)

\(\Rightarrow a=b=c=d\)

\(M=\frac{2a-b}{c+b}+\frac{2b-c}{a+d}+\frac{2c-d}{a+b}+\frac{2d-a}{b+c}=\frac{2a-a}{a+a}+\frac{2b-b}{b+b}+\frac{2c-c}{c+c}+\frac{2d-d}{d+d}=\frac{1}{2}.4=2\)

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

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

Với \(a+b+c\ne0\) áp dụng tính chất dãy tỉ số bằng nhau ta có:

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

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

Khi đó:

\(P=\dfrac{2a}{a}+\dfrac{2b}{b}+\dfrac{2c}{c}=2+2+2=6\)

Áp dụng Bdt cosi 3 số dương ta có"

\(a^3+b^3+c^3\ge3\sqrt[3]{a^3b^3c^3}=3abc\)

Dấu = khi a=b=c

Đpcm

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

=> \(\frac{a+b-c}{c}+2=\frac{b+c-a}{a}+2=\frac{c+a-b}{b}+2\)

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

Nếu a + b + c = 0

=> a + b = -c

b + c = -a

a + c = -b

Khi đó P = \(\left(1+\frac{b}{a}\right)\left(1+\frac{c}{b}\right)\left(1+\frac{a}{c}\right)=\frac{a+b}{a}.\frac{b+c}{b}.\frac{a+c}{c}=\frac{-c}{a}.\frac{-a}{b}.\frac{-b}{c}=\frac{-abc}{abc}=-1\)

Nếu a + b + c \(\ne\)0

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

=> a = b = c

Khi đó P \(\left(1+\frac{b}{a}\right)\left(1+\frac{c}{b}\right)\left(1+\frac{a}{c}\right)=\left(1+1\right)\left(1+1\right)\left(1+1\right)=2.2.2=8\)

Vậy khi a + b + c = 0 thì P = -1

khi a + b + c  \(\ne\)0 thì P = 8