Vì \(a+b+c=2016\Rightarrow a=2016-\left(b+c\right);b=2016-\left(a+c\right);c=2016-\left(a+b\right)\)

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

           \(S=\frac{2016-\left(b+c\right)}{b+c}+\frac{2016-\left(a+c\right)}{a+c}+\frac{2016-\left(a+b\right)}{a+b}\)

           \(S=\frac{2016}{b+c}-1+\frac{2016}{a+c}-1+\frac{2016}{a+b}-1\)

           \(S=2016.\left(\frac{1}{b+c}+\frac{1}{a+c}+\frac{1}{a+b}\right)-3\)

           \(S=2016.\frac{1}{2016}-3\)

          \(S=-2\)

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

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

\(\Leftrightarrow\frac{a+b}{ab}=\frac{c-a-b-c}{c\left(a+b+c\right)}\)

\(\Leftrightarrow\frac{a+b}{ab}=\frac{-\left(a+b\right)}{c\left(a+b+c\right)}\)

\(\Leftrightarrow c\left(a+b\right)\left(a+b+c\right)=-ab\left(a+b\right)\)

\(\Leftrightarrow\left(a+b\right)\left(ac+bc+c^2\right)+ab\left(a+b\right)=0\)

\(\Leftrightarrow\left(a+b\right)\left(ac+bc+c^2+ab\right)=0\)

\(\Leftrightarrow\left(a+b\right)\left[c\left(a+c\right)+b\left(a+c\right)\right]=0\)

\(\Leftrightarrow\left(a+b\right)\left(a+c\right)\left(b+c\right)=0\)

TH1 : \(a+b=0\Leftrightarrow a=-b\)

\(M=\left(-b^{15}+b^{15}\right)\left(b^4+c^4\right)\left(c^{2016}+a^{2016}\right)\)

\(M=0\left(b^4+c^4\right)\left(c^{2016}+a^{2016}\right)=0\)

TH2 : \(b+c=0\Leftrightarrow b=-c\)

Đến đây tịt :) bác nào biết giải tiếp giúp Nghị Hồng Vân Anh

đề cho a,b trái dấu rồi nên có một trường hợp thôi nha Trần Thanh Phương, cảm ơn bạn

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

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

Thay các dữ liệu mà đề bài cho vào S,ta có:

\(S=2016.\frac{1}{90}-3=\frac{97}{5}\)

Vậy S=97/5

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

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

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

\(S+3=\frac{2014.1}{2014}=1\Rightarrow S=1-3=-2\)

ta có: \(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}=\frac{1}{7}\)

\(\Rightarrow14.\left(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}\right)=14.\frac{1}{7}\)

\(\Rightarrow\frac{14}{a+b}+\frac{14}{b+c}+\frac{14}{c+a}=2\)

mà a+b+c =14

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

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

\(\Rightarrow3+\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}=2\)

\(\Rightarrow A=\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}=2-3\)

\(\Rightarrow A=-1\)

CHÚC BN HỌC TỐT!!!!!!

tìm x y z biết

\(\sqrt{2016.x^2+4}+\sqrt{2017y^2+9}=9-\sqrt{2019z^2+25}\)

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