Question

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

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

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

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

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

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

\(D=2+2+2\)

\(D=6\)

 

Answer 2

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

=>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

 ^...^  ^_^

Answer 3

a/b+c=b/a+c=c/a+b=a+b+c/b+c+a+c+a+b=a+b+c/2a+2b+2c=a+b+c/2(a+b+c)=1/2

⇒a+c=2b⇒a+c=2b

⇒a+b=2c⇒a+b=2c

D=b+c/a+a+c/b+a+b/c

D=2a/a=2b/b=2c/c

D=2+2+2

D=6