Question

cho a, b, c thỏa mãn

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

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

Answer 1

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

Nếu \(a+b+c\ne0\) thì tỉ số trên bằng 1 . Từ đó , ta có : \(a=b=c\)

\(B=\left(1+\dfrac{b}{a}\right)\left(1+\dfrac{c}{b}\right)\left(1+\dfrac{a}{b}\right)=2.2.2=8\)

P/s : Tớ làm bừa :D

Answer 2

Ta có:

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

\(\Rightarrow\dfrac{a+b-c}{a}=1\Rightarrow a+b-c=a\Rightarrow b-c=0\Rightarrow b=c\left(1\right)\)

\(\dfrac{b+c-a}{c}=1\Rightarrow b+c-a=c\Rightarrow b-a=0\Rightarrow b=a\left(2\right)\)

Từ (1) và (2) suy ra: a = b = c

Suy ra:

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