Question

Cho a,b,c > 0 thỏa mãn a+b+c = 2019 và \(\frac{1}{a+b}\)+\(\frac{1}{b+c}\)+\(\frac{1}{c+a}\)= 10 . Tính giá trị của P = \(\frac{a}{b+c}\)+\(\frac{b}{c+a}\)+\(\frac{c}{a+b}\)

Answer 1

Ta có : \(P=\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\)

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

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

\(\Rightarrow P+3=\left(a+b+c\right).\frac{1}{b+c}+\left(a+b+c\right).\frac{1}{c+a}+\left(a+b+c\right).\frac{1}{a+b}\)

\(\Rightarrow P+3=\left(a+b+c\right).\left(\frac{1}{b+c}+\frac{1}{c+a}+\frac{1}{a+b}\right)\)

\(\Rightarrow P+3=2019.10\)

\(\Rightarrow P+3=20190\)

\(\Rightarrow P=20190-3\)

\(\Rightarrow P=20187\)

Vậy P = 20187