a) Ta có : \(\frac{a+b}{c}=\frac{b+c}{a}=\frac{c+a}{b}\Leftrightarrow\frac{a+b}{c}+1=\frac{b+c}{a}+1=\frac{c+a}{b}+1\)

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

• TH1: Nếu a + b + c = 0 \(\Rightarrow P=\frac{a+b}{b}.\frac{b+c}{c}.\frac{a+c}{a}=\frac{-c}{b}.\frac{-a}{c}.\frac{-b}{a}=\frac{-\left(abc\right)}{abc}=-1\)
• TH2 : Nếu \(a+b+c\ne0\) \(\Rightarrow a=b=c\)

\(\Rightarrow P=\left(1+1\right)\left(1+1\right)\left(1+1\right)=8\)

b) Đề bài sai ^^

xét a + b + c = 0 khi đó a + b = -c ; b + c = -a ; a + c = -b

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

xét a + b + c \(\ne\)0 . thì \(\frac{a+b}{c}=\frac{b+c}{a}=\frac{c+a}{b}=\frac{2\left(a+b+c\right)}{a+b+c}=2\)

\(\Rightarrow a+b=2c;b+c=2a\)\(\Rightarrow a-c=2\left(c-a\right)\)\(\Rightarrow a=c\)( loại vì a khác c )

Vậy A = -1

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

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

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

\(\Rightarrow a=b=c\)

\(\Rightarrow\frac{b}{a}=1;\frac{a}{c}=1;\frac{c}{b}=1\)

\(\Rightarrow B=\left(1+\frac{b}{a}\right)\left(1+\frac{a}{c}\right)\left(1+\frac{c}{b}\right)=\left(1+1\right)\left(1+1\right)\left(1+1\right)=8\)

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

=> a= b =c 

=> P = (1+1) ( 1+1)(1+1) = 2.2.2 =8

cảm ơn

Ta có:

a+b-c/c = b+c-a/a = c+a-b/b

=>a+b-c/c + 2 = b+c-a/a +2 = c+a-b/b +2

=>a+b-c/c  + 2c/c =b+c-a/a +2a/a = c+a-b/b +2/b

=>a+b+c/c = a+b+c/a =a+b+c/b

* Nếu a+b+c=0 thì a= 0-b-c= -(b+c)

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

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

Thay a= -(b+c) ; b=-(a+c);c=-(b+a) vào B ta được

B=(1+b/a)(1+a/c)(1+c/b)=(a/a + b/a )(c/c +a/c)(b/b+c/b)=(a+b)/a * (a+c)/c * (c+b)/b

                                                                                =(-c)/a * (-b)/c * (-a)/b =-1

* Nếu  a+b+c\(\ne\)0 thì a=b=c

Khi đó

B=(1+b/a)(1+a/c)(1+c/b)=(1+1)(1+1)(1+1)=2*2*2=8

Vậy B=-1 hoặc B=8

nhớ k nha bạn

B=1 hoặc B=8 nha!

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