Question

[★ღTrúc Lyღ★]

Cho a+b+c= 2016 và 1/ a+b + 1/a+c + 1/b+c= 1/2016

Tính S= a/b+c + b/a+c + c/a+b

Mai nộp bài r ạ....giúp mìk vs

Answer 1

[༂d̲̅a̲̅r̲̅k̲̅☬s̲̅t̲̅a̲̅r̲̅➻]

Vì a + c = 2016 -> a = 2016 - [ b + c] ; b = 2016 - [ a + c] ; c = 2016 - [ a - b]

Ta có: S = a/ b + c   +  b/ a + c   + c/a + b

S = 2016 - [ b + c] + 2016 - [ a + c] + 2016 - [ a + b]

S = 2016/ b + c - 1 + 2016/a + c - 1 + 2016/a + b

S = 2016.[ 1/b + c   + 1/a + c  + 1/a + b] - 3

S = 2016. 1/2016 - 3

S = - 2

Answer 2

[Chu Công Đức]

Từ \(a+b+c=2016\) và \(\frac{1}{a+b}+\frac{1}{a+c}+\frac{1}{b+c}=\frac{1}{2016}\)

\(\Rightarrow\left(a+b+c\right)\left(\frac{1}{a+b}+\frac{1}{a+c}+\frac{1}{b+c}\right)=2016.\frac{1}{2016}\)

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

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

\(\Rightarrow1+\frac{c}{a+b}+1+\frac{b}{a+c}+1+\frac{a}{b+c}=1\)

\(\Rightarrow\frac{c}{a+b}+\frac{b}{a+c}+\frac{a}{b+c}=-2\)

hay \(P=-2\)