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4 tháng 3 2017

Ta có :

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

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

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

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

\(=2009.\frac{1}{7}=287\)

\(\Rightarrow S=287-3=284\)

22 tháng 2 2017

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

\(\Leftrightarrow\frac{1}{a}+\frac{1}{b}+\frac{1}{c}-\frac{1}{a+b+c}=0\)

\(\Leftrightarrow\frac{a+b}{ab}+\frac{a+b+c-c}{c\left(a+b+c\right)}=0\)

\(\Leftrightarrow\frac{a+b}{ab}+\frac{a+b}{c\left(a+b+c\right)}=0\)

\(\Leftrightarrow\left(a+b\right)\left(\frac{1}{ab}+\frac{1}{c\left(a+b+c\right)}\right)=0\)

\(\Leftrightarrow\left(a+b\right)\left(\frac{ca+cb+c^2+ab}{abc\left(a+b+c\right)}\right)=0\)

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

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

\(\Rightarrow a+b=0\Rightarrow a=-b\Rightarrow a^{2009}=-b^{2009}\)

\(\frac{1}{a^{2009}}+\frac{1}{b^{2009}}+\frac{1}{c^{2009}}=\frac{1}{c^{2009}}\) (1)

\(\frac{1}{a^{2009}+b^{2009}+c^{2009}}=\frac{1}{c^{2009}}\) (2)

Từ (1) và (2) \(\Rightarrow\frac{1}{a^{2009}}+\frac{1}{b^{2009}}+\frac{1}{c^{2009}}=\frac{1}{a^{2009}+b^{2009}+c^{2009}}\) (đpcm)

16 tháng 1 2018

=> (a+b+c).(1/a+b + 1/b+c  +1/c+a) = 2017/90

=> a+b+c/a+b + a+b+c/b+c + a+b+c/c+a = 2017/90

=> 1 + c/a+b + 1 + a/b+c + 1 + b/c+a = 2017/90

=> a/b+c + b/c+a  +c/a+b = 2017/90 - 3 = 1747/90

Vậy S = 1747/90

Tk mk nha

8 tháng 1 2018

a+b+c = 2010 => a+b=2010-c ; b+c=2010-a ; c+a=2010-b

=> S = a/2010-a + b/2010-b + c/2010-c = 2010/2010-a - 1 + 2010/2010-b -1 + 2010/2010-c - 1

= 2010/b+c - 1 + 2010/c+a - 1 + 2010/a+b - 1

= 2010.(1/b+c + 1/c+a + 1/a+b) - 3 

= 2010.1/3 - 3 = 667

Vậy S = 667

Tk mk nha

8 tháng 1 2018

Ta có: \(\left(a+b+c\right)\left(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}\right)=2010\cdot\frac{1}{3}\)

\(\Rightarrow\frac{a+b+c}{a+b}+\frac{a+b+c}{b+c}+\frac{a+b+c}{c+a}=\frac{2010}{3}\)

\(\Rightarrow1+\frac{c}{a+b}+1+\frac{a}{b+c}+1+\frac{b}{c+a}=\frac{2010}{3}\)

\(\Rightarrow S+3=\frac{2010}{3}\)

\(\Rightarrow S=\frac{2010}{3}-3=\frac{2001}{3}=667\)

25 tháng 4 2021

Ta có S + 4 = \(\left(\frac{a}{b+c+d}+1\right)+\left(\frac{b}{c+d+a}+1\right)+\left(\frac{c}{a+b+d}+1\right)+\left(\frac{d}{a+b+c}+1\right)\)

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

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

\(=4000.\frac{1}{40}=100\)(a + b + c + d = 4000 ; \(\frac{1}{b+c+d}+\frac{1}{a+c+d}+\frac{1}{a+b+d}+\frac{1}{a+b+c}=\frac{1}{40}\))

=> S = 100 - 4 = 96

30 tháng 3 2018

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!!!!!!

3 tháng 12 2018

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

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

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

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

\(=2007.\frac{1}{90}=\frac{223}{10}\Rightarrow S=\frac{223}{10}-3=\frac{193}{10}\)

3 tháng 3 2020

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

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

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

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

\(=2007.\frac{1}{90}=\frac{223}{10}\)

\(=>S=\frac{223}{10}-\frac{30}{10}=\frac{193}{10}\)

6 tháng 5 2019

ta có \(\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{b+c}\)

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

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

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

rồi còn lại thay vào nha bn

6 tháng 5 2019

\(\left(a+b+c\right)\left(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}\right)=2019\cdot\frac{1}{2019}\)

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

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

\(\Leftrightarrow S=-2\)

8 tháng 1 2018

Ta có: \(\left(a+b+c\right)\left(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}\right)=2017\cdot\frac{1}{90}\)

\(\Rightarrow\frac{a+b+c}{a+b}+\frac{a+b+c}{b+c}+\frac{a+b+c}{c+a}=\frac{2017}{90}\)

\(\Rightarrow1+\frac{c}{a+b}+1+\frac{a}{b+c}+1+\frac{b}{c+a}=\frac{2017}{90}\)

\(\Rightarrow A+3=\frac{2017}{90}\)

\(\Rightarrow S=\frac{2017}{90}-3=\frac{1747}{90}\)

8 tháng 1 2018

từ giả thiết, ta có 

\(\frac{1}{2017-a}+\frac{1}{2017-b}+\frac{1}{2017-c}=\frac{1}{90}\)

Mà \(S=\frac{a}{2017-a}+\frac{b}{2017-b}+\frac{c}{2017-c}=-3+\frac{2017}{2017-a}+\frac{2017}{2017-b}+\frac{2017}{2017-c}\)

=-3+\(2017\left(\frac{1}{2017-a}+\frac{1}{2017-b}+\frac{1}{2017-c}\right)=-3+\frac{2017}{90}=\frac{1747}{90}\)

vậy ...

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