Cho a + b + c = 2018 và \(\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a}=\dfrac{1}{10}\). Tính \(S=\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}\)
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AH
Akai Haruma
Giáo viên
30 tháng 4 2023
Lời giải:
\((a+b+c)(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{a+c})=\frac{a}{a+b}+\frac{a}{b+c}+\frac{a}{a+c}+\frac{b}{a+b}+\frac{b}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}+\frac{c}{b+c}+\frac{c}{a+c}\)
$\Leftrightarrow 2018.\frac{1}{2018}=\frac{a+b}{a+b}+\frac{b+c}{b+c}+\frac{c+a}{c+a}+\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}$
$\Leftrightarrow 1=1+1+1+S$
$S=1-1-1-1=-2$
17 tháng 5 2017
Sửa đề:
\(S=\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}\)
\(=\left(\dfrac{a}{b+c}+1\right)+\left(\dfrac{b}{c+a}+1\right)+\left(\dfrac{c}{a+b}+1\right)-3\)
\(=\dfrac{a+b+c}{b+c}+\dfrac{a+b+c}{c+a}+\dfrac{a+b+c}{a+b}-3\)
\(=\left(a+b+c\right)\left(\dfrac{1}{b+c}+\dfrac{1}{c+a}+\dfrac{1}{a+b}\right)-3\)
\(=2001.\dfrac{1}{10}-3\)
\(=200,1-3=197,1\)
Vậy S = 197,1
9 tháng 7 2017
\(\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{a+c}=\dfrac{1}{10}\)
\(\Rightarrow2017\left(\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{a+c}\right)=\dfrac{2017}{10}\)
\(\Rightarrow\dfrac{2017}{a+b}+\dfrac{2017}{b+c}+\dfrac{2017}{a+c}=201,7\)
\(\Rightarrow\dfrac{a+b+c}{a+b}+\dfrac{a+b+c}{b+c}+\dfrac{a+b+c}{a+c}=201,7\)
\(\Rightarrow\dfrac{a+b}{a+b}+\dfrac{c}{a+b}+\dfrac{b+c}{b+c}+\dfrac{a}{b+c}+\dfrac{a+c}{a+c}+\dfrac{b}{a+c}=201,7\)
\(\Rightarrow1+\dfrac{c}{a+b}+1+\dfrac{a}{b+c}+1+\dfrac{b}{a+c}=201,7\)
\(\Rightarrow3+\dfrac{c}{a+b}+\dfrac{a}{b+c}+\dfrac{b}{a+c}=201,7\)
\(\Rightarrow\dfrac{c}{a+b}+\dfrac{a}{b+c}+\dfrac{b}{a+c}=198,7\)
AT
9 tháng 7 2017
Ta có: \(\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a}=\dfrac{1}{10}\)
\(=>2017\left(\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a}\right)=\dfrac{2017}{10}\)
\(=>\dfrac{2017}{a+b}+\dfrac{2017}{b+c}+\dfrac{2017}{c+a}=201,7\)
Mà 2017 = a+b+c nên ta có:
\(=>\dfrac{a+b+c}{a+b}+\dfrac{a+b+c}{b+c}+\dfrac{a+b+c}{c+a}=201,7\)
\(=>1+\dfrac{c}{a+b}+1+\dfrac{a}{b+c}+1+\dfrac{b}{a+c}=201,7\)
\(=>\dfrac{a}{b+c}+\dfrac{b}{a+c}+\dfrac{c}{a+b}=201,7-3=198,7\)
CHÚC BẠN HỌC TỐT....
E
5 tháng 10 2018
Vì a + b + c = 2018
\(\Rightarrow\left\{{}\begin{matrix}b+c=2018-a\\c+a=2018-b\\a+b=2018-c\end{matrix}\right.\)
Ta có: \(P=\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}=\dfrac{a}{2018-a}+\dfrac{b}{2018-b}+\dfrac{c}{2018-c}\)
\(P+3=\left(\dfrac{a}{2018-a}+1\right)+\left(\dfrac{b}{2018-b}+1\right)+\left(\dfrac{c}{2018-c}+1\right)=\dfrac{2018}{b+c}+\dfrac{2018}{c+a}+\dfrac{2018}{a+b}=2018\left(\dfrac{1}{b+c}+\dfrac{1}{c+a}+\dfrac{1}{a+c}\right)=2018.\dfrac{2017}{2018}=2017\Rightarrow P=2014\)
6 tháng 10 2018
Ta có : \(P=\dfrac{a}{b+c}+\dfrac{b}{a+c}+\dfrac{c}{b+a}\)
\(\Rightarrow3+P=\left(\dfrac{a}{b+c}+1\right)+\left(\dfrac{b}{a+c}+1\right)+\left(\dfrac{c}{a+b}+1\right)\)
\(\Rightarrow3+P=\dfrac{a+b+c}{b+c}+\dfrac{a+b+c}{a+c}+\dfrac{a +b+c}{a+b}\)
\(\Rightarrow3+P=\left(a+b+c\right).\left(\dfrac{1}{b+c}+\dfrac{1}{a+c}+\dfrac{1}{a+b}\right)\)
Mà \(a+b+c=2018;\) \(\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a}=\dfrac{2017}{2018}\) \(\left(a,b\in R\right)\)
\(\Rightarrow3+P=2018.\dfrac{2017}{2018}\)
\(\Rightarrow3+P=2017\)
\(\Rightarrow P=2014\)
Vậy \(P=2014\)