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Ta có: \(S=\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\)
\(3+S=\left(1+\frac{a}{b+c}\right)+\left(1+\frac{b}{c+a}\right)+\left(1+\frac{c}{a+b}\right)\)
\(=\frac{a+b+c}{b+c}+\frac{a+b+c}{c+a}+\frac{a+b+c}{a+b}\)
\(=\left(a+b+c\right)\left(\frac{1}{b+c}+\frac{1}{c+a}+\frac{1}{a+b}\right)\)
\(=2007.\frac{1}{90}=\frac{223}{10}\Rightarrow S=\frac{223}{10}-3=\frac{193}{10}\)
\(S=\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\)
\(=>S+3=\frac{a}{b+c}+\frac{b+c}{b+c}+\frac{b}{c+a}+\frac{c+a}{c+a}+\frac{c}{a+b}+\frac{a+b}{a+b}\)
\(=\frac{a+b+c}{b+c}+\frac{a+b+c}{c+a}+\frac{a+b+c}{a+b}\)
\(=\left(a+b+c\right)\left(\frac{1}{b+c}+\frac{1}{c+a}+\frac{c}{a+b}\right)\)
\(=2007.\frac{1}{90}=\frac{223}{10}\)
\(=>S=\frac{223}{10}-\frac{30}{10}=\frac{193}{10}\)
Ta có: \(\left(a+b+c\right)\left(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}\right)=2017\cdot\frac{1}{90}\)
\(\Rightarrow\frac{a+b+c}{a+b}+\frac{a+b+c}{b+c}+\frac{a+b+c}{c+a}=\frac{2017}{90}\)
\(\Rightarrow1+\frac{c}{a+b}+1+\frac{a}{b+c}+1+\frac{b}{c+a}=\frac{2017}{90}\)
\(\Rightarrow A+3=\frac{2017}{90}\)
\(\Rightarrow S=\frac{2017}{90}-3=\frac{1747}{90}\)
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 ...
^_^
Ta có :
\(A+3=\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}+3\)
\(=\left(\frac{a}{b+c}+1\right)+\left(\frac{b}{a+c}+1\right)+\left(\frac{c}{a+b}+1\right)\)
\(=\frac{a+b+c}{b+c}+\frac{a+b+c}{a+c}+\frac{a+b+c}{a+b}\)
\(=\left(a+b+c\right)\left(\frac{1}{b+c}+\frac{1}{a+c}+\frac{1}{a+b}\right)\)
\(=2017.\frac{1}{2017}=1\)
\(\Rightarrow A=1-3=-2\)
=> (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
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