\(\frac{a+b-c}{c}=\frac{b+c-a}{a}=\frac{c+a-b}{b}\)
\(\Rightarrow\frac{a+b-c}{c}+1=\frac{b+c-a}{a}+1=\frac{c+a-b}{b}+1\)
\(\Rightarrow\frac{a+b}{c}=\frac{b+c}{a}=\frac{c+a}{b}\)
+)Nếu a+b+c=0\(\Rightarrow a+b=-c;b+c=-a;c+a=-b\)
\(\Rightarrow B=\frac{a+b}{a}.\frac{c+a}{c}.\frac{b+c}{b}=\frac{-c}{a}.\frac{-b}{c}.\frac{-a}{b}=\frac{-\left(abc\right)}{abc}=-1\)
Nếu \(a+b+ c\ne0\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có
\(\frac{a+b}{c}=\frac{b+c}{a}=\frac{c+a}{b}=\frac{2\left(a+b+c\right)}{a+b+c}=2\)
\(\Rightarrow a+b=2c\)
\(b+ c=2a\)
\(c+a=2b\)
\(\Rightarrow B=\frac{2c}{a}.\frac{2b}{c}.\frac{2a}{b}=2.2.2=8\)
a+b-c/c=b+c-a/a=c+a-b/b
=>a+b-1=b+c-1=c+a-1
=>a+b=b+c=c+a
Vì a+b=b+c
=>a=b+c-b
=>a=c
Vì b+c=c+a
=>b=c+a-c
=>b=a
Mà a=c
=>a=b=c
Ta có:B=(1+b/a).(1+a/c).(1+c/b)
=>B=(1+b/b).(1+a/a).(1+c/c)
=>B=(1+1).(1+1).(1+1)
=>B=2.2.2
=>B=8
Vậy B=8
Hok tốt!
⇒a+b−c/c +1=b+c−a/a +1=c+a−b/b +1
⇒a+bc =b+ca =c+ab
Áp dụng tính chất dãy tỉ số bằng nhau ta có
a+bc =b+ca =c+ab =2(a+b+c)/a+b+c =2
⇒a+b=2c
b+c=2a
c+a=2b
=>B=2c/a*2b/c*2a/b=2*2*2=8