Question

cho a+b+c=2016 và \(\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a}=\dfrac{1}{4}\)

Tính \(A=\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}\)

Answer 1

theo bài ra ta có:

\(\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a}=\dfrac{`1}{4}\)

\(\Rightarrow\left(a+b+c\right)\left(\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a}\right)=\dfrac{1}{4}\left(a+b+c\right)\)

\(\Rightarrow\dfrac{a+b+c}{a+b}+\dfrac{a+b+c}{b+c}+\dfrac{a+b+c}{c+a}=\dfrac{a+b+c}{4}\)

\(\Rightarrow1+\dfrac{c}{a+b}+\dfrac{a}{b+c}+1+\dfrac{b}{c+a}+1=\dfrac{2016}{4}\)

\(\Rightarrow\left(1+1+1\right)+\left(\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}\right)=504\)

\(\Rightarrow3+\left(\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}\right)=504\)

\(\Rightarrow\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}=504-3\)

\(\Rightarrow\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}=501\)

vậy \(\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}=501\)

Answer 2

(a+b+c)(1/a+b+1/b+c+1/c+a)=(a+b+c)/4

(a+b+c)/(a+b)+(a+b+c)/(b+c)+(a+b+c)/(c+a)=(a+b+c)/4

=> 1+c/(a+b)+1+a/(b+c)+1+b/(c+a)=2016/4

<=>c/(a+b)+a/(b+c)+b/(c+a)+3=504

=> A=a/(b+c)+b/(c+a)+c/(a+b)=504-3=501

Answer 3

a.b.c