Cho a + b + c = 2017
Và \(\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{a+c}=\dfrac{1}{10}\)
Tính S = \(\dfrac{a}{b+c}+\dfrac{b}{a+c}+\dfrac{c}{a+b}\)
K
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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
AH
Akai Haruma
Giáo viên
30 tháng 4 2023
Lời giải:
\((a+b+c)(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{a+c})=\frac{a}{a+b}+\frac{a}{b+c}+\frac{a}{a+c}+\frac{b}{a+b}+\frac{b}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}+\frac{c}{b+c}+\frac{c}{a+c}\)
$\Leftrightarrow 2018.\frac{1}{2018}=\frac{a+b}{a+b}+\frac{b+c}{b+c}+\frac{c+a}{c+a}+\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}$
$\Leftrightarrow 1=1+1+1+S$
$S=1-1-1-1=-2$
\(\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{a+c}=\dfrac{1}{10}\)
\(\Rightarrow2017\left(\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{a+c}\right)=\dfrac{2017}{10}\)
\(\Rightarrow\dfrac{2017}{a+b}+\dfrac{2017}{b+c}+\dfrac{2017}{a+c}=201,7\)
\(\Rightarrow\dfrac{a+b+c}{a+b}+\dfrac{a+b+c}{b+c}+\dfrac{a+b+c}{a+c}=201,7\)
\(\Rightarrow\dfrac{a+b}{a+b}+\dfrac{c}{a+b}+\dfrac{b+c}{b+c}+\dfrac{a}{b+c}+\dfrac{a+c}{a+c}+\dfrac{b}{a+c}=201,7\)
\(\Rightarrow1+\dfrac{c}{a+b}+1+\dfrac{a}{b+c}+1+\dfrac{b}{a+c}=201,7\)
\(\Rightarrow3+\dfrac{c}{a+b}+\dfrac{a}{b+c}+\dfrac{b}{a+c}=201,7\)
\(\Rightarrow\dfrac{c}{a+b}+\dfrac{a}{b+c}+\dfrac{b}{a+c}=198,7\)
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:
\(=>\dfrac{a+b+c}{a+b}+\dfrac{a+b+c}{b+c}+\dfrac{a+b+c}{c+a}=201,7\)
\(=>1+\dfrac{c}{a+b}+1+\dfrac{a}{b+c}+1+\dfrac{b}{a+c}=201,7\)
\(=>\dfrac{a}{b+c}+\dfrac{b}{a+c}+\dfrac{c}{a+b}=201,7-3=198,7\)
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