Question

Cho \(\frac{a}{b+c}\)=\(\frac{b}{c+a}=\frac{c}{a+b}\)Tính giá trị của biểu thức B=\(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\)

Answer 1

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

=> b+c=2a; c+a=2b; a+b=2c

=> \(B=\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}=\frac{a}{2a}+\frac{b}{2b}+\frac{c}{2c}=\frac{1}{2}+\frac{1}{2}+\frac{1}{2}\)

=> \(B=\frac{3}{2}\)

Answer 2

Áp dụng tính chất dãy tỉ số bằng nhau ta được:

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

Ta có \(\frac{a}{b+c}=\frac{1}{2}\Leftrightarrow b+c=2a\)

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

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

Lại có  :

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

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

\(B=\frac{1}{2}+\frac{1}{2}+\frac{1}{2}\)

\(B=\frac{3}{2}\)