Question

[Đoàn Phong]

Cho 3 số a, b, c khác nhau và khác 0, thoả mãn:

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

Tính: \(D=\frac{b+c}{a}+\frac{a+c}{b}=\frac{a+b}{c}\)

Answer 1

[Võ Đông Anh Tuấn]

\(\frac{a}{b+c}=\frac{b}{a+c}=\frac{c}{a+b}=\frac{a+b+c}{b+c+a+c+a+b}=\frac{a+b+c}{2a+2b+2c}=\frac{a+b+c}{2\left(a+b+c\right)}=\frac{1}{2}\)\(\Rightarrow b+c=2a\)

\(\Rightarrow a+c=2b\)

\(\Rightarrow a+b=2c\)

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

\(D=\frac{2a}{a}=\frac{2b}{b}=\frac{2c}{c}\)

\(D=2+2+2\)

\(D=6\)

Answer 2

[Trịnh Thị Như Quỳnh]

\(\frac{a}{b+c}=\frac{b}{a+c}=\frac{c}{a+b}=\frac{a+b+c}{b+c+a+c+a+b}=\frac{a+b+c}{2a+2b+2c}=\frac{a+b+c}{2\left(a+b+c\right)}=\frac{1}{2}\)

=>b+c=2a

=>a+c=2b

=>a+b=2c

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

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

\(D=2+2+2\)

D=6

Vậy D=6

 ^...^  ^_^