Ta có :
\(\dfrac{a+b-c}{c}=\dfrac{b+c-a}{a}=\dfrac{c+a-b}{b}\)
\(\Leftrightarrow\dfrac{a+b-c}{c}+2=\dfrac{b+c-a}{a}+2=\dfrac{c+a-b}{b}+2\)
\(\Leftrightarrow\dfrac{a+b-c+2c}{c}=\dfrac{b+c-a+2a}{a}=\dfrac{c+a-b+2b}{b}\)
\(\Leftrightarrow\dfrac{a+b+c}{c}=\dfrac{b+c+a}{a}=\dfrac{c+a+b}{b}\)
\(\Leftrightarrow\dfrac{1}{c}=\dfrac{1}{a}=\dfrac{1}{b}\)
\(\Rightarrow a=b=c\)
\(B=\left(1+\dfrac{b}{a}\right)\left(1+\dfrac{a}{c}\right)\left(1+\dfrac{b}{c}\right)\)
\(=\left(1+1\right)\left(1+1\right)\left(1+1\right)\\ =2\cdot2\cdot2\\ =8\)
Ta có : \(\dfrac{a+b-c}{c}=\dfrac{b+c-a}{a}=\dfrac{c+a-b}{b}\)
\(\Rightarrow\dfrac{a+b-c}{c}+2=\dfrac{b+c-a}{a}+2=\dfrac{c+a-b}{b}+2\)
\(=\dfrac{a+b+c}{c}=\dfrac{a+b+c}{a}=\dfrac{a+b+c}{b}\)
\(\Rightarrow a=b=c\)
\(\Rightarrow B=\left(1+\dfrac{b}{a}\right)\left(1+\dfrac{a}{c}\right)\left(1+\dfrac{b}{c}\right)=\left(1+1\right)\left(1+1\right)\left(1+1\right)\)
\(=2.2.2=8\)
Áp dụng tính chất của dãy tỉ số = nhau được:\(\dfrac{a+b-c}{c}=\dfrac{b+c-a}{a}=\dfrac{c+a-b}{b}=\dfrac{a+b-c+b+c-a+c+a-b}{c+a+b}=\dfrac{a+b+c}{a+b+c}=1\)
\(\Rightarrow\dfrac{a+b-c}{c}=1\Rightarrow a+b=2c\) (1)
\(\dfrac{b+c-a}{a}=1\Rightarrow b+c=2a\) (2)
\(\dfrac{c+a-b}{b}=1\Rightarrow c+a=2b\) (3)
Thay (1), (2), (3) vào B ta được :
B = \(\left(1+\dfrac{b}{a}\right)\cdot\left(1+\dfrac{a}{c}\right)\cdot\left(1+\dfrac{c}{b}\right)\)
= \(\left(\dfrac{a}{a}+\dfrac{b}{a}\right)\cdot\left(\dfrac{c}{c}+\dfrac{a}{c}\right)\cdot\left(\dfrac{b}{b}+\dfrac{c}{b}\right)\)
= \(\left(\dfrac{a+b}{a}\right)\left(\dfrac{c+a}{c}\right)\left(\dfrac{b+c}{b}\right)\)
= \(\dfrac{2c}{a}\cdot\dfrac{2b}{c}\cdot\dfrac{2a}{b}=\dfrac{2c\cdot2b\cdot2a}{a\cdot c\cdot b}=8\)
Vậy B = 8
