a+b+c bằng mấy

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

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

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

\(=2010.\frac{1}{3}=670\)

\(\Rightarrow S=670-3=667\)

\(a+b+c=2016\Rightarrow\left\{{}\begin{matrix}a=2016-\left(b+c\right)\\b=2016-\left(c+a\right)\\c=2016-\left(a+b\right)\end{matrix}\right.\)

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

\(\Rightarrow S=2016.\dfrac{1}{90}-3\)

\(\Rightarrow S=\dfrac{97}{2}\)

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

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

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

Nếu a + b + c = 0

=> a + b = -c

b + c = -a

a + c = -b

Khi đó P = \(\left(1+\frac{b}{a}\right)\left(1+\frac{c}{b}\right)\left(1+\frac{a}{c}\right)=\frac{a+b}{a}.\frac{b+c}{b}.\frac{a+c}{c}=\frac{-c}{a}.\frac{-a}{b}.\frac{-b}{c}=\frac{-abc}{abc}=-1\)

Nếu a + b + c \(\ne\)0

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

=> a = b = c

Khi đó P \(\left(1+\frac{b}{a}\right)\left(1+\frac{c}{b}\right)\left(1+\frac{a}{c}\right)=\left(1+1\right)\left(1+1\right)\left(1+1\right)=2.2.2=8\)

Vậy khi a + b + c = 0 thì P = -1

khi a + b + c  \(\ne\)0 thì P = 8

\(\frac{a+b}{c}=\frac{b+c}{a}=\frac{c+a}{b}\Rightarrow\frac{a+b+c}{c}=\frac{a+b+c}{a}=\frac{a+b+c}{b}\)    (cộng 3 vế với 1)

TH1: \(a+b+c=0\)

Khi đó: \(M=\left(\frac{a+b}{b}\right)\left(\frac{b+c}{c}\right)\left(\frac{c+a}{a}\right)=\frac{-c}{b}.\frac{-a}{c}.\frac{-b}{a}=-1\)

TH2: \(a=b=c\) (ko thỏa mãn a,b,c đôi 1 khác nhau)

Vây M = -1

Chúc bạn học tốt.

ta có: \(\frac{a+b}{c}=\frac{b+c}{a}=\frac{c+a}{b}=\frac{a+b+b+c+c+a}{c+a+b}=\frac{2.\left(a+b+c\right)}{a+b+c}.\)

Nếu \(a+b+c\ne0\)thì \(\frac{2.\left(a+b+c\right)}{a+b+c}=2\)

=> a + b = 2c

b+c = 2a

=> a-c = 2.(c-a)

=> c=a ( trái với đề bài)

=> a + b +c = 0

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