Question

[Nguyến Gia Hân]

Cho a + b + c = 2019 và \(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}=\frac{1}{2019}\) 

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

Answer 1

[karipham]

ta có \(\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{b+c}\)

=\(\frac{a}{b+c}+1+\frac{b}{a+c}+1+\frac{c}{b+c}+1-3\)

=\(\frac{a+b+c}{b+c}+\frac{a+b+c}{a+c}+\frac{a+b+c}{a+b}-3\)

=\(\left(a+b+c\right)\left(\frac{1}{c+b}+\frac{1}{a+c}+\frac{1}{a+b}\right)-3\)

rồi còn lại thay vào nha bn

Answer 2

[zZz Cool Kid_new zZz]

\(\left(a+b+c\right)\left(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}\right)=2019\cdot\frac{1}{2019}\)

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

\(\Leftrightarrow\left(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\right)+3=1\)

\(\Leftrightarrow S=-2\)