Question

cho a+b+c = 2007 và \(\frac{1}{a+b}\)=\(\frac{1}{b+c}\)=\(\frac{1}{c+a}\)=\(\frac{1}{90}\). Tính S=\(\frac{a}{b+c}\)=\(\frac{b}{c+a}\)=\(\frac{c}{a+b}\)

Answer 1

ta có \(S=\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{b+a}\)

=>\(S+3=3+\left(\dfrac{a}{b+c}+\dfrac{c}{b+a}+\dfrac{b}{c+a}\right)\)

hay \(S+3=\left(\dfrac{a}{b+c}+1\right)+\left(\dfrac{b}{c+a}+1\right)+\left(\dfrac{c}{b+a}+1\right)\)

=>\(S+3=\dfrac{a+b+c}{b+c}+\dfrac{a+b+c}{a+c}+\dfrac{a+b+c}{b+a}\)

=>\(S+3=a+b+c\left(\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a}\right)\)

=>\(S+3=2007\cdot\dfrac{1}{90}\)

=>\(S+3=\dfrac{2017}{90}\)

=>S=\(\dfrac{1747}{90}\)