Question

Giả sử a,b,c là các số thỏa mãn \(a+b+c=259\) và \(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{a+c}=15\) .

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

Answer 1

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

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

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

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

\(\Rightarrow Q=259.15-3=3882\)

Answer 2

(a+b+c)(1/a+b + 1/b+c + 1/a+c)=259.15

(a+b+c).(1/a+b) + (a+b+c).(1/b+c) + (a+b+c).(1/a+c)=259.15

a+b+c/a+b  +  a+b+c/b+c   +   a+b+c/a+c=259.15

(1  +  c/a+b)     +        (1  +  a/b+c)    +    (1  +   b/a+c)=259.15

3+ (c/a+b  +   a/b+c   +   b/a+c)=259.15

c/a+b   +   a/b+c     +       b/a+c= 259.15-3

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