Cho \(a+b+c=2012\) và \(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}=\frac{1}{503}\). Tính: \(S=\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\)
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N
10 tháng 12 2018
\(S=\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}\)
\(S+3=\left(1+\frac{a}{b+c}\right)+\left(1+\frac{b}{a+c}\right)+\left(1+\frac{c}{a+b}\right)\)
\(S+3=\frac{a+b+c}{b+c}+\frac{a+b+c}{a+c}+\frac{a+b+c}{a+b}=\left(a+b+c\right).\left(\frac{1}{b+c}+\frac{1}{a+c}+\frac{1}{a+b}\right)\)
\(S+3=\frac{2014.1}{2014}=1\Rightarrow S=1-3=-2\)
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
S
8 tháng 1 2018
Ta có: \(\left(a+b+c\right)\left(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}\right)=2010\cdot\frac{1}{3}\)
\(\Rightarrow\frac{a+b+c}{a+b}+\frac{a+b+c}{b+c}+\frac{a+b+c}{c+a}=\frac{2010}{3}\)
\(\Rightarrow1+\frac{c}{a+b}+1+\frac{a}{b+c}+1+\frac{b}{c+a}=\frac{2010}{3}\)
\(\Rightarrow S+3=\frac{2010}{3}\)
\(\Rightarrow S=\frac{2010}{3}-3=\frac{2001}{3}=667\)
XO
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
T
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
\(S=\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\)
\(=>S+3=\frac{a}{b+c}+\frac{b+c}{b+c}+\frac{b}{c+a}+\frac{c+a}{c+a}+\frac{c}{a+b}+\frac{a+b}{a+b}\)
\(=\frac{a+b+c}{b+c}+\frac{a+b+c}{c+a}+\frac{a+b+c}{a+b}\)
\(=\left(a+b+c\right)\left(\frac{1}{b+c}+\frac{1}{c+a}+\frac{c}{a+b}\right)\)
\(=2007.\frac{1}{90}=\frac{223}{10}\)
\(=>S=\frac{223}{10}-\frac{30}{10}=\frac{193}{10}\)
X
25 tháng 1 2019
Đặt : \(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}=P\)
\(\Rightarrow\left(a+b+c\right).P=\frac{1}{2019}.2019\)
\(\Rightarrow1+\frac{c}{a+b}+1+\frac{a}{b+c}+1+\frac{b}{c+a}=\frac{6057}{2019}+\frac{\left(-4038\right)}{2019}\)
\(\Rightarrow3+\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}=3+\left(-2\right)\)
\(\Rightarrow\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}=-2\)
K
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
\(\left(a+b+c\right)\left(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}\right)=2019\cdot\frac{1}{2019}\)
\(\Leftrightarrow\frac{a+b+c}{a+b}+\frac{a+b+c}{b+c}+\frac{a+b+c}{c+a}=1\)
\(\Leftrightarrow\left(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\right)+3=1\)
\(\Leftrightarrow S=-2\)