Câu hỏi của NGô Văn Trường

Question

Cho a,b,c$\ne$0 thỏa mãn:

$\frac{b+c-a}{a}=\frac{a+c-b}{b}=\frac{a+b-c}{c}$

Tính \(\left(1+\frac{b}{a}\r...

Nguyễn Thị Thùy Dương · Accepted Answer

$\frac{b+c-a}{a}+\frac{2a}{a}=\frac{a+c-b}{b}+\frac{2b}{b}=\frac{a+b-c}{c}+\frac{2c}{c}$

$\frac{a+b+c}{a}=\frac{a+b+c}{b}=\frac{a+b+c}{c}$

=> a=b=c

A=(1+1)(1+1)(1+1) = 2.2.2 =8

Câu trả lời:

$\frac{b+c-a}{a}+\frac{2a}{a}=\frac{a+c-b}{b}+\frac{2b}{b}=\frac{a+b-c}{c}+\frac{2c}{c}$

$\frac{a+b+c}{a}=\frac{a+b+c}{b}=\frac{a+b+c}{c}$

=> a=b=c

A=(1+1)(1+1)(1+1) = 2.2.2 =8