Question

cho a+b+c=4034 và \(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}=\frac{1}{2}\)

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

ai giải đc mình tick cho

Answer 1

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

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

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

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

\(\Rightarrow S=4034.\frac{1}{2}-3=2014\)

Vậy S=2014

Answer 2

key ban di chi gium n he khong k luon nhe

Answer 3

mình ko hiểu